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name
stringlengths 2
112
| description
stringlengths 29
13k
| source
int64 1
7
| difficulty
int64 0
25
| solution
stringlengths 7
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| language
stringclasses 4
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|---|---|---|---|---|---|
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(true){
final int N = sc.nextInt();
final int K = sc.nextInt();
if(N == 0 && K == 0){
break;
}
int[] drinks = new int[K];
for(int i = 0; i < K; i++){
drinks[i] = sc.nextInt();
}
boolean flag = true;
for(int i = 0; i < N; i++){
for(int j = 0; j < K; j++){
drinks[j] -= sc.nextInt();
if(drinks[j] < 0){
flag = false;
}
}
}
System.out.println(flag ? "Yes" : "No");
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
int N,K;
int S[101];
int B;
int main(){
while(cin >> N >> K,N != 0){
for(int i = 0;i < K; i++){
cin >> S[i];
}
for(int i = 0; i < N; i++){
for(int j = 0; j < K; j++){
cin >> B;
S[j] -= B;
}
}
bool f = true;
for(int i = 0; i < K; i++){
if(S[i] < 0){
cout << "No" << endl;
f = false;
break;
}
}
if(f){
cout << "Yes" << endl;
}
}
return 0;
}
| CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner stdIn = new Scanner(System.in);
while(true) {
int n = stdIn.nextInt();
int k = stdIn.nextInt();
if(n + k == 0) break;
int[] ary = new int[k];
for(int r = 0; r < k; r++) {
ary[r] = stdIn.nextInt();
}
boolean flag = true;
for(int r = 0; r < n; r++) {
for(int c = 0; c < k; c++) {
int blood = stdIn.nextInt();
ary[c] -= blood;
if(ary[c] < 0) flag = false;
}
}
if(flag) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
using namespace std;
#define MAX_N 120
int S[MAX_N];
int sum[MAX_N];
int n, k, a;
int main() {
while (true) {
for (int i = 0; i < n; i++) {
sum[i] = 0;
}
cin >> n >> k;
if (n == 0 && k == 0) { break; }
for (int i = 0; i < k; i++) {
cin >> S[i];
}
bool ok = true;
for (int i = 0; i < n; i++) {
for (int j = 0; j < k; j++) {
cin >> a;
S[j] -= a;
if (S[j] < 0) { ok = false; }
}
}
if (ok == true) { cout << "Yes" << endl; }
else { cout << "No" << endl; }
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]){
Scanner in = new Scanner(System.in);
while(true){
int n=in.nextInt(), k=in.nextInt();
if(n==0 && k==0) break;
int ss[] = new int[k];
for(int i=0; i<k; i++){
ss[i] = in.nextInt();
}
for(int i=0; i<n; i++){
for(int j=0; j<k; j++){
ss[j] -= in.nextInt();
}
}
int i=0;
for(i=0; i<k; i++){
if(ss[i] < 0) break;
}
System.out.println(i==k?"Yes":"No");
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <sstream>
using namespace std;
int main(){
int N,K,zaiko[128],a,flag;
for(;cin>>N>>K,N||K;){
flag=0;
for(int i=0;i<K;i++)cin>>zaiko[i];
for(int i=0;i<N;i++){
for(int j=0;j<K;j++){
cin>>a; zaiko[j]-=a;
if(zaiko[j]<0)flag=1;
}
}
puts(flag?"No":"Yes");
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main(void){
int n,k,s[100],b;
while(cin >> n >> k,n|k){
for(int i=0;i<k;i++)cin >> s[i];
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
cin >> b;
s[j]-=b;
}
}
bool fg=false;
for(int i=0;i<k;i++){
if(s[i]<0){
cout << "No" << endl;
fg=true;
break;
}
}
if(!fg)cout << "Yes" << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <algorithm>
#include <cassert>
#include <cctype>
#include <complex>
#include <cstdio>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
int N,K,S[100],B[100][100];
int main(){
while(cin>>N>>K){
if(!N)return 0;
for(int i=0;i<K;i++)cin>>S[i];
for(int i=0;i<N;i++)for(int j=0;j<K;j++)cin>>B[i][j];
int f=1;
for(int i=0;i<K;i++){
int s=0;
for(int j=0;j<N;j++)s+=B[j][i];
if(s>S[i]){f=0;break;}
}
if(f)cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.*;
public class Main {
public static void main(String[] args) {
try(Scanner scn = new Scanner(System.in)) {
int n, k;
while((n = scn.nextInt()) > 0 | (k = scn.nextInt()) > 0) {
int[] s = new int[k];
for(int i = 0; i < k; i++) {
s[i] = scn.nextInt();
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < k; j++) {
s[j] -= scn.nextInt();
}
}
Arrays.sort(s);
System.out.println(s[0] < 0 ? "No" : "Yes");
}
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int main(){
int n,k;
int s[1000];
int b;
while(true){
memset(s,0,sizeof(s));
cin >> n >> k;
if(!n && !k) break;
for(int i=0;i<k;i++) cin >> s[i];
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
cin >> b;
s[j] -= b;
}
}
bool d = true;
for(int i=0;i<k;i++){
if(s[i] < 0) d = false;
}
if(!d) cout << "No" << endl;
else cout << "Yes" << endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
using namespace std;
int main(void){
int n,k;
int s[150];
int b[150][150];
int a,f;
for( ; ; ){
cin >> n >> k;
if(!n && !k) break;
f = 0;
for(int i = 0;i < k; i++) cin >> s[i];
for(int i = 0;i < n; i++){
for(int j = 0;j < k; j++){
cin >> a;
s[j] -= a;
}
}
for(int i = 0;i < k; i++){
if(s[i] < 0) f = 1;
}
cout << (f? "No" : "Yes") << endl;
}
}
| CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <stdio.h>
int main(){
while(1){
int n,k;
scanf("%d%d", &n, &k);
if(n==0&&k==0) break;
int max[k];
int dat[k];
int i,j;
for(i=0;i<k;i++) scanf("%d", max+i);
for(i=0;i<k;i++) dat[i] = 0;
int ok=1;
for(i=0;i<n;i++){
for(j=0;j<k;j++){
int v;
scanf("%d", &v);
dat[j]+=v;
if(dat[j] > max[j]) ok = 0;
}
}
if(ok) puts("Yes");
else puts("No");
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <vector>
#include <string>
using namespace std;
int main(void)
{
int n, k, tmp;
vector <int> blood;
vector <string> result;
while (true) {
cin >> n >> k;
if (n == 0 && k == 0)
break;
blood.reserve(k);
for (int i = 0; i <= n; i++)
for (int j = 0; j < k; j++) {
cin >> tmp;
if (i == 0)
blood[j] = tmp;
else
blood[j] -= tmp;
}
tmp = 1;
for (int i = 0; i < k; i++)
if (blood[i] < 0) {
tmp = 0;
break;
}
result.push_back(tmp ? "Yes" : "No");
}
for (int i = 0; i < result.size(); i++)
cout << result[i] << endl;
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <algorithm>
using namespace std;
int main(void)
{
int N, K;
while (cin >> N >> K, N && K){
int blood[128];
int eat[128];
fill (blood, blood + 128, 0);
fill (eat, eat + 128, 0);
for (int i = 0; i < K; i++){
cin >> blood[i];
}
int tmp;
for (int i = 0; i < N; i++){
for (int j = 0; j < K; j++){
cin >> tmp;
eat[j] += tmp;
}
}
bool flag = true;
for (int i = 0; i < K; i++){
if (blood[i] < eat[i]) flag = false;
}
if (flag) cout << "Yes" << endl;
else cout << "No" << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
using namespace std;
int main()
{
int n, k;
while (cin >> n >> k, n || k){
int s[101], b[101];
bool f = true;
for (int i = 0; i < k; i++){
cin >> s[i];
}
for (int i = 0; i < n; i++){
for (int j = 0; j < k; j++){
cin >> b[j];
s[j] -= b[j];
}
}
for (int i = 0; i < k; i++){
if (s[i] < 0){
f = false;
}
}
if (f){
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
}
return (0);
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Scanner;
class Main{
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
while(true){
int n = scan.nextInt();
int k = scan.nextInt();
if(n==0&&k==0){
break;
}
int[] s = new int[k];
boolean flag=true;
for(int i=0;i<k;i++){
s[i]=scan.nextInt();
}
int[] adder = new int[k];
for(int j=0;j<n;j++){
for(int i=0;i<k;i++){
adder[i]+=scan.nextInt();
}
}
for(int i=0;i<k;i++){
if(adder[i]>s[i]){
flag=false;
}
}
if(flag){
System.out.println("Yes");
}
else{
System.out.println("No");
}
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.awt.geom.Point2D;
import java.util.*;
public class Main {
Scanner in = new Scanner(System.in);
public static void main(String[] args) {
new Main();
}
public Main() {
while(in.hasNext())new AOJ1019().doIt();
}
class AOJ1019{
boolean solve(){
int[] blood = new int[k];
for(int i=0;i<k;i++)blood[i] = in.nextInt();
// System.out.println(Arrays.toString(blood));
int person[][] = new int[n][k];
for(int i=0;i<n;i++)for(int s=0;s<k;s++)person[i][s] = in.nextInt();
boolean sw = true;
for(int i=0;i<k;i++){
int sum = 0;
for(int s=0;s<n;s++)sum+=person[s][i];
// System.out.println(i+" "+sum+" "+blood[i]);
if(sum>blood[i])return false;
}
return true;
}
int n,k;
void doIt(){
n = in.nextInt();
k = in.nextInt();
if(n+k==0)return;
System.out.println(solve()? "Yes":"No");
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
using namespace std;
int main(){
int N,K,i,j,S[100],B,a;
while(a=1,cin>>N>>K,N||K){
for(i=K;i--;)cin>>S[i];
for(i=N;i--;)for(j=K;j--;)cin>>B,S[j]-=B,S[j]<0?a=0:0;
cout<<(a?"Yes":"No")<<endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(true){
String nk = sc.nextLine();
String[] nka = nk.split(" ");
int n = Integer.parseInt(nka[0]);
int k = Integer.parseInt(nka[1]);
if(n == 0 && k == 0) break;
int[][] all = new int[n+1][k];
boolean jud = true;
for(int i = 0 ; i < all.length ; i++){
String str = sc.nextLine();
String[] stra = str.split(" ");
for(int j = 0 ; j < stra.length ; j++){
all[i][j] = Integer.parseInt(stra[j]);
}
}
for(int i = 1 ; i < all.length ;i++){
for(int j = 0 ; j < all[i].length ; j++){
all[0][j] -= all[i][j];
}
}
for(int i = 0 ; i < all[0].length ;i++){
if(all[0][i] < 0) jud = false;
}
if(jud){
System.out.println("Yes");
} else {
System.out.println("No");
}
}
sc.close();
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
#include<cstdio>
using namespace std;
int main(){
int n,k;
int i,j;
while(cin>>n>>k){
if(n==0 && k==0){break;}
int s[100];
int b[100][100];
for(i=0;i<k;i++){
cin>>s[i];
}
for(i=0;i<n;i++){
for(j=0;j<k;j++){
cin>>b[i][j];
}
}
for(i=1;i<n;i++){
for(j=0;j<k;j++){
b[0][j]+=b[i][j];
}
}
bool futo=true;
for(j=0;j<k;j++){
if(s[j]-b[0][j]>=0){
}else{
futo=false;
}
}
if(futo)
cout<<"Yes";
else cout<<"No";
cout<<endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.io.*;
public class Main {
public static void main(String[] args) {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
try {
String line;
while(true){
/* input */
line = br.readLine();
int n = Integer.parseInt(line.split(" ")[0]), k = Integer.parseInt(line.split(" ")[1]);
if(n==0&&k==0) break;
int[] s = new int[k];
int[][] b = new int[n][k];
line = br.readLine();
String[] str = line.split(" ");
for(int i=0;i<k;i++){
s[i] = Integer.parseInt(str[i]);
}
for(int i=0;i<n;i++){
line = br.readLine();
str = line.split(" ");
for(int j=0;j<k;j++){
b[i][j] = Integer.parseInt(str[j]);
}
}
/* processing */
boolean f = true;
for(int i=0;i<k;i++){
int sum = 0;
for(int j=0;j<n;j++){
sum += b[j][i];
}
if(sum>s[i]){
f = false;
break;
}
if(!f) break;
}
/* output */
if(f){
System.out.println("Yes");
} else {
System.out.println("No");
}
}
} catch (IOException e) {
e.printStackTrace();
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<cstdio>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int N,K;
int S[101];
int main(){
while(1){
scanf("%d %d",&N,&K);
if(!N && !K) break;
for(int i=0;i<K;i++){
scanf("%d",&S[i]);
}
bool f = false;
for(int i=0;i<N;i++){
for(int j=0;j<K;j++){
int b;
scanf("%d",&b);
S[j]-=b;
if(S[j]<0) f=true;
}
}
if(f) printf("No\n");
else printf("Yes\n");
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
using namespace std;
static const double EPS = 1e-8;
typedef long long ll;
int s[100000];
int main()
{
int N, K;
while (scanf("%d%d", &N, &K), N | K)
{
for (int i = 0; i < K; ++i)
scanf("%d", s + i);
bool yes = true;
while (N--)
{
for (int i = 0; i < K; ++i)
{
int b;
scanf("%d", &b);
s[i] -= b;
if (s[i] < 0)
yes = false;
}
}
if (yes)
puts("Yes");
else
puts("No");
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<stdio.h>
int main(){
int N,K;
int S[100];
int B;
int count;
int h=1;
int i,j;
scanf("%d %d",&N,&K);
while((N!=0)&&(K!=0)){
h=1;
for(i=0;i<K;i++){
scanf("%d",&S[i]);
}
for(i=0;i<N;i++){
for(j=0;j<K;j++){
scanf("%d",&B);
if(h==1){
S[j]=S[j]-B;
if(S[j]<0){
h=0;
printf("No\n");
}
}
}
}
if(h==1){
printf("Yes\n");
}
scanf("%d %d",&N,&K);
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(true){
int[] s = new int[100000];
boolean ans = true;
int N = sc.nextInt();
int K = sc.nextInt();
if(N == 0 && K == 0) break;
for(int k = 0; k < K; k++){
s[k] = sc.nextInt();
}
for(int n = 0; n < N; n++){
for(int k = 0; k < K; k++){
s[k] -= sc.nextInt();
}
}
for(int k = 0; k < K; k++){
if(s[k] < 0){
ans = false;
break;
}
}
if(ans == true) System.out.println("Yes");
else System.out.println("No");
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <utility>
using namespace std;
int N,K,S[100];
int main() {
while(true) {
scanf("%d %d", &N, &K);
if (N == 0 && K == 0) break;
for (int i = 0; i < K; ++i) {
scanf("%d", S+i);
}
for (int i = 0; i < N; ++i) {
for (int j = 0; j < K; ++j) {
int B;
scanf("%d", &B);
S[j]-=B;
}
}
cout << (*min_element(S, S+K) >= 0 ? "Yes" : "No") << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = "";
StringTokenizer st;
StringBuilder sb = new StringBuilder();
int[] stock = new int[100];
int[] needs = new int[100];
next: while ((line = br.readLine()) != null && !line.isEmpty()) {
int n = Integer.parseInt(line.substring(0, line.indexOf(' ')));
int k = Integer.parseInt(line.substring(line.indexOf(' ') + 1));
if ((n | k) == 0) {
break;
}
st = new StringTokenizer(br.readLine());
for (int i = 0; i < k; i++) {
stock[i] = Integer.parseInt(st.nextToken());
needs[i] ^= needs[i];
}
for (int i = 0; i < n; i++) {
st = new StringTokenizer(br.readLine());
for (int j = 0; j < k; j++) {
needs[j] += Integer.parseInt(st.nextToken());
}
}
for (int i = 0; i < k; i++) {
if (stock[i] < needs[i]) {
sb.append("No\n");
continue next;
}
}
sb.append("Yes\n");
}
System.out.print(sb.toString());
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;
int sum[100], need[100];
int main() {
int n, k;
while (cin >> n >> k, n) {
rep(i, k)scanf("%d", &sum[i]);
memset(need, 0, sizeof(need));
rep(i, n)rep(j, k) {
int a; cin >> a;
need[j] += a;
}
bool flag = true;
rep(i, k) {
if (need[i] > sum[i]) {
puts("No");
flag = false;
break;
}
}
if (flag)puts("Yes");
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <vector>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int N, K;
while (cin >> N >> K && (N || K)) {
vector<int> S(K);
for (int i=0; i<K; ++i) cin >> S[i];
int t;
bool ok = true;
for (int i=0; i<N; ++i) {
for (int j=0; j<K; ++j) {
cin >> t;
if (ok) {
S[j] -= t;
if (S[j] < 0) {
ok = false;
}
}
}
}
cout << (ok ? "Yes" : "No") << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 |
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
public class Main {
int INF = 1 << 28;
void run() {
Scanner sc = new Scanner(System.in);
for(;;) {
int n = sc.nextInt();
int k = sc.nextInt();
if( (n|k) == 0 ) break;
int[] br = new int[k];
boolean flg = true;
for(int i=0;i<k;i++) br[i] = sc.nextInt();
for(int i=0;i<n;i++) for(int j=0;j<k;j++) {
br[j] -= sc.nextInt();
if( br[j] < 0 ) flg = false;
}
System.out.println((flg? "Yes": "No"));
}
}
public static void main(String[] args) {
new Main().run();
}
void debug(Object... os) {
System.err.println(Arrays.deepToString(os));
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <algorithm>
#include <iostream>
using namespace std;
int main()
{
int n, k;
while(cin >> n >> k, n&&k) {
int s[100], b[100][100];
for(int i = 0; i < k; i++) {
cin >> s[i];
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < k; j++) {
cin >> b[i][j];
}
}
bool f = false;
for(int i = 0; i < n; i++) {
for(int j = 0; j < k; j++) {
s[j] -= b[i][j];
if(s[j] < 0) {
f = true;
break;
}
}
}
cout << (f ? "No" : "Yes") << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <limits.h>
#include <queue>
#include <algorithm>
#include <map>
#include <cstring>
#define dprintf(s,...) printf("%s:%d" s,__func__,__LINE__,__VA_ARGS__)
using namespace std;
typedef pair<int,int> P;
typedef long long int ll;
int N,K;
int S[101],B[101][101];
bool solve(){
for(int i=0;i<K;i++){
int t=0;
for(int j=0;j<N;j++)
t+=B[j][i];
if(t>S[i]) return false;
}
return true;
}
int main(){
while(1){
cin>>N>>K;
if(N==0 && K==0) break;
for(int i=0;i<K;i++)
cin>>S[i];
for(int i=0;i<N;i++)
for(int j=0;j<K;j++)
cin>>B[i][j];
if(solve())cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
while(1){
int n,k;
cin >> n >> k;
if(n==0) break;
vector<int> blood(k);
for(int i=0; i<k; i++) cin >> blood[i];
for(int i=0; i<n; i++){
for(int j=0; j<k; j++){
int a;
cin >> a;
blood[j] -= a;
}
}
bool success = true;
for(int i=0; i<k; i++){
if(blood[i] < 0) success = false;
}
if(success){
cout << "Yes" << endl;
}else{
cout << "No" << endl;
}
}
return 0;
}
| CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
using namespace std;
int main() {
int n, k;
while (cin >> n >> k,n,k) {
int s[100];
for (int i = 0;i < k;i++) {cin >> s[i];}
int b[100] = {};
for (int i = 0;i < n;i++) {
for (int j = 0;j < k;j++) {
int a;
cin >> a;
b[j] += a;
}
}
int flg = 0;
for (int i = 0;i < k;i++) {
if (b[i] > s[i]) {
flg = 1;
break;
}
}
if (flg == 0)cout << "Yes" << endl;
else cout << "No" << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n,k,s[110],sum[110],a,i,j;
while(scanf("%d%d",&n,&k)&&(n||k))
{
memset(sum,0,sizeof(sum));
for(j=1;j<=k;j++) scanf("%d",&s[j]);
for(j=1;j<=n;j++)
{
for(i=1;i<=k;i++)
{
scanf("%d",&a);
sum[i]+=a;
}
}
bool ok=1;
for(i=1;i<=k;i++) if(s[i]<sum[i]) ok=0;
if(ok) printf("Yes\n");
else printf("No\n");
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;
int main()
{
for(;;){
int n, k;
cin >> n >> k;
if(n == 0)
return 0;
vector<int> blood(k);
for(int i=0; i<k; ++i)
cin >> blood[i];
for(int i=0; i<n; ++i){
for(int j=0; j<k; ++j){
int tmp;
cin >> tmp;
blood[j] -= tmp;
}
}
if(*(min_element(blood.begin(), blood.end())) < 0)
cout << "No" << endl;
else
cout << "Yes" << endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int[] s = new int[100];
while (n != 0 || k != 0) {
Arrays.fill(s, 0);
for (int i = 0; i < k; i++) {
s[i] = in.nextInt();
}
boolean isYes = true;
for (int i = 0; i < n; i++) {
for (int j = 0; j < k; j++) {
if (isYes) {
s[j] -= in.nextInt();
isYes = s[j] >= 0;
} else {
in.next();
}
}
}
System.out.println(isYes ? "Yes" : "No");
n = in.nextInt();
k = in.nextInt();
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
static int N, K;
static int[] s;
static int[][] b;
public static void main(String[] args) {
while(read()){
solve();
}
}
static boolean read(){
N = sc.nextInt(); K = sc.nextInt();
if( N == 0 && K == 0 )return false;
s = new int[K];
b = new int[N][K];
for(int i = 0; i < s.length; i++){
s[i] = sc.nextInt();
}
for(int i = 0; i < N; i++){
for(int j = 0; j < K; j++){
b[i][j] = sc.nextInt();
}
}
return true;
}
static void solve(){
for(int i = 0; i < N; i++){
for(int j = 0; j < K; j++){
s[j] -= b[i][j];
}
}
boolean flag = true;
for(int i = 0; i < s.length; i++){
if( s[i] < 0 )flag = false;
}
if(flag){
System.out.println("Yes");
}else{
System.out.println("No");
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <string>
using namespace std;
int main() {
int N, K;
while (cin >> N >> K) {
if (N == 0 && K == 0) break;
int S[K];
for (int i = 0; i < K; i++) {
cin >> S[i];
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < K; j++) {
int b; cin >> b;
S[j] -= b;
}
}
string ans = "Yes";
for (int i = 0; i < K; i++) {
if (S[i] < 0) {
ans = "No";
}
}
cout << ans << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
using namespace std;
int main(void)
{
int N, K;
while(cin >> N >> K){
if(!N) break;
int S[100]={0};
for(int i=0; i<K; i++){
cin >> S[i];
}
bool f=false;
for(int i=0; i<N; i++){
int B[100]={0};
for(int j=0; j<K; j++){
cin >> B[j];
S[j] -= B[j];
if(S[j]<0) {
f = true;
}
}
}
if(f) cout << "No\n";
else cout << "Yes\n";
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
using namespace std;
int n, m, x, s[100];
int main() {
while(cin >> n >> m, n + m) {
for(int i = 0; i < m; i++) cin >> s[i];
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) cin >> x, s[j] -= x;
}
int f = 1;
for(int i = 0; i < m; i++) {
if(s[i] < 0) f = 0;
}
cout << (f ? "Yes" : "No") << endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
//1019
static class VampirishNight {
void print_solution() {
for(;;) {
int n = sc.nextInt();
int k = sc.nextInt();
if(n == 0 && k == 0) break;
int[] s = new int[k];
for(int i=0; i<k; i++) {s[i] = sc.nextInt();}
String ans = "Yes";
for(int i=0; i<n; i++) {
for(int j=0; j<k; j++) {
s[j] -= sc.nextInt();
if(s[j] < 0) ans = "No";
}
}
System.out.println(ans);
}
}
}
public static void main(String[] args) {
VampirishNight vn = new VampirishNight();
vn.print_solution();
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <set>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,j) FOR(i,0,j)
#define mp std::make_pair
const int INF = 1 << 24;
const int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1}, dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
typedef unsigned long long ull;
typedef std::pair<int,int> P;
int main(){
int N, K;
while(std::cin >> N >> K, N){
int S[100];
REP(i, K){
std::cin >> S[i];
}
bool canHave = true;
REP(i, N){
REP(j, K){
int b;
std::cin >> b;
S[j] -= b;
if(S[j] < 0){canHave = false;}
}
}
if(canHave){puts("Yes");}
else{puts("No");}
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.io.*;
import java.util.ArrayList;
import java.util.List;
// main
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//String linebr.readLine()) != null;
while (true) {
List<Integer> amanda;
amanda = getInt(br);
int num = amanda.get(0);
int kind = amanda.get(1);
if(num == 0 && kind == 0) {
break;
}
List<Integer> goukei = getInt(br);
List<Integer> init = new ArrayList<Integer>();
for(int i=0; i<kind; i++) {
init.add(0);
}
for(int i=0; i<num; i++) {
int r = 0;
for(int op : getInt(br)){
init.set(r,init.get(r) + op);
r++;
}
}
String ans = "Yes";
for(int i=0; i<kind; i++) {
if(goukei.get(i) < init.get(i)) {
ans = "No";
}
}
System.out.println(ans);
}
}
static List<Integer> getInt(BufferedReader bk) throws IOException {
String[] a = bk.readLine().split(" ");
List<Integer> k = new ArrayList<Integer>();
for(String string : a) {
int b = Integer.parseInt(string);
k.add(b);
}
return k;
}
}
| JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <vector>
using namespace std;
int main(){
int n, k;
while(cin >> n >> k, n+k){
vector<int>s;
for(int i=0; i < k; i++){
int num;
cin >> num;
s.push_back(num);
}
for(int i=0; i < n; i++){
for(int j=0; j < k; j++){
int num;
cin >> num;
s[j]-=num;
}
}
bool isPossible = true;
for(int i=0; i < s.size(); i++){
if(s[i] < 0) isPossible = false;
}
if(isPossible) cout << "Yes" << endl;
else cout << "No" << endl;
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner stdIn = new Scanner(System.in);
while(true){
int n = stdIn.nextInt();
int k = stdIn.nextInt();
if(n==0&&k==0){
return;
}
int[] have = new int[k];
for(int i=0;i<k;i++){
have[i] = stdIn.nextInt();
}
boolean f = true;
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
have[j] -= stdIn.nextInt();
if(have[j]<0){
f = false;
}
}
}
if(f){
System.out.println("Yes");
}
else{
System.out.println("No");
}
}
}
} | JAVA |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include "bits/stdc++.h"
using namespace std;
//#define int long long
#define DEBUG 0
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(a) (a).begin(),(a).end()
#define dump(o) if(DEBUG){cerr<<#o<<" "<<o<<endl;}
#define dumpc(o) if(DEBUG){cerr<<#o; for(auto &e:(o))cerr<<" "<<e;cerr<<endl;}
using ll = long long; using ull = unsigned long long; using pii = pair<int, int>;
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
static const int MOD = 1e9 + 7;
signed main() {
for (int N, K; cin >> N >> K&&N;) {
vector<int> S(K);
rep(i, 0, K)cin >> S[i];
rep(i, 0, N)rep(j, 0, K) {
int B; cin >> B;
S[j] -= B;
}
if (*min_element(all(S)) >= 0) {
cout << "Yes" << endl;
}
else cout << "No" << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
#include<string>
using namespace std;
int main(){
int n,k,b,s[101],c[101];
while(true){
cin>>n>>k;
if(n==0&&k==0)
break;
for(int i=0;i<k;i++){
cin>>s[i];
c[i]=0;
}
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
cin>>b;
c[j]+=b;
}
}
bool t=true;
for(int i=0;i<k;i++){
if(s[i]<c[i])
t=false;
}
if(t==true)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | while 1:
n, k = map(int, raw_input().split())
if n == 0: break
l = map(int, raw_input().split())
for _ in range(n):
m = map(int, raw_input().split())
for i in range(k):
l[i] -= m[i]
print "No" if min(l) < 0 else "Yes" | PYTHON |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
int main()
{
int N, K;
while (scanf("%d %d ", &N, &K) == 2) {
if (N == 0 && K == 0) {
break;
}
// printf("(N, K) = (%d, %d)\n", N, K);
vector<int> S(K);
for (int i = 0; i < K; i++) {
int s;
scanf("%d ", &s);
S[i] = s;
}
bool flag = true;
for (int i = 0; i < N; i++) {
for (int j = 0; j < K; j++) {
int b;
scanf("%d ", &b);
S[j] -= b;
if (S[j] < 0) {
flag = false;
}
}
}
if (flag) {
printf("Yes\n");
} else {
printf("No\n");
}
}
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include <iostream>
using namespace std;
int main() {
while(1) {
int n,k;
cin >> n >> k;
if(n == 0 && k == 0) break;
int a,flag = 0;
int K[101];
for(int i=0;i<k;i++)
cin >> K[i];
for(int i=0;i<n;i++) {
for(int j=0;j<k;j++) {
cin >> a;
K[j] -= a;
if(K[j] < 0) flag =1;
}
}
if(flag == 1) cout << "No" << endl;
else cout << "Yes" << endl;
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
int n,k;
int i,l;
int sum;
int flag;
int s[100];
int b[100][100];
for(;;){
cin>>n>>k;
if(n==0&&k==0) break;
memset(s,0,sizeof(s));
memset(b,0,sizeof(b));
flag=0;
for(i=0;i<k;i++) cin>>s[i];
for(i=0;i<n;i++){
for(l=0;l<k;l++){
cin>>b[i][l];
}
}
for(i=0;i<k;i++){
sum=0;
for(l=0;l<n;l++){
sum+=b[l][i];
}
if(sum>s[i]){
puts("No");
flag++;
break;
}
}
if(flag==0) puts("Yes");
}
return 0;
} | CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<iostream>
#include<vector>
using namespace std;
bool solve(int n,int k){
vector<int> s(k);
vector<int> v(k,0);
for(int i = 0;i < k;i++)cin >> s[i];
for(int i = 0;i < n;i++)for(int j = 0;j < k;j++){
int g;
cin >> g;
v[j] +=g;
}
bool q = true;
for(int i = 0;i < k;i++){
if(v[i] > s[i]){
return false;
}
}
return true;
}
int main(){
vector<bool>a;
while(1){
int n,k;
cin >> n >> k;
if(n == 0)break;
a.push_back(solve(n,k));
}
for(int i = 0;i < a.size();i++){
if(a[i] == true){
cout << "Yes" <<endl;
}
else{
cout << "No" <<endl;
}
}
return 0;
}
| CPP |
p00605 Vampirish Night | There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
You, cook of the family, are looking inside a fridge to prepare dinner. Your first task is to write a program that judges if all family members' dinner can be prepared using blood in the fridge.
Constraints
* Judge data includes at most 100 data sets.
* 1 ≤ N ≤ 100
* 1 ≤ K ≤ 100
* 0 ≤ Si ≤ 100000
* 0 ≤ Bij ≤ 1000
Input
Input file consists of a number of data sets. One data set is given in following format:
N K
S1 S2 ... SK
B11 B12 ... B1K
B21 B22 ... B2K
:
BN1 BN2 ... BNK
N and K indicate the number of family members and the number of blood types respectively.
Si is an integer that indicates the amount of blood of the i-th blood type that is in a fridge.
Bij is an integer that indicates the amount of blood of the j-th blood type that the i-th vampire uses.
The end of input is indicated by a case where N = K = 0. You should print nothing for this data set.
Output
For each set, print "Yes" if everyone's dinner can be prepared using blood in a fridge, "No" otherwise (without quotes).
Example
Input
2 3
5 4 5
1 2 3
3 2 1
3 5
1 2 3 4 5
0 1 0 1 2
0 1 1 2 2
1 0 3 1 1
0 0
Output
Yes
No | 7 | 0 | #include<stdio.h>
int s[101],n,k,v,o,i;int main(){for(;scanf("%d%d",&n,&k)&&n&&k;puts(o?"No":"Yes")){for(i=k;i--;scanf("%d",s+i))o=0;for(;n--;)for(i=k;i--;o|=(s[i]-=v)<0)scanf("%d",&v);}}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
#define N 15
int imax, jmax, solution, digit[256], low[256];
char word[N][16];
bool ok[10];
void found(void){
for(int i=0; i<imax; i++){
if(strlen(word[i])>1&&digit[word[i][strlen(word[i])-1]]==0)return;
}
++solution;
}
void solve(int sum){
static int i=0, j=0, carry;
int c, d;
c = word[i][j];
if(i < imax-1) {
i++;
if((d = digit[c]) < 0) {
for(d = low[c]; d <= 9; d++)
if(ok[d]) {
digit[c] = d; ok[d] = false;
solve(sum + d); ok[d] = true;
}
digit[c] = -1;
} else solve(sum + d);
i--;
} else {
j++; i=0; d = sum % 10; carry = sum / 10;
if(digit[c] == d) {
if(j <= jmax) solve(carry);
else if(carry == 0) found();
} else if(digit[c] < 0 && ok[d] && d >= low[c]) {
digit[c] = d; ok[d] = false;
if(j <= jmax) solve(carry);
else if(carry == 0) found();
digit[c] = -1; ok[d] = true;
}
j--; i = imax-1;
}
}
int main(){
int k, c;
char buffer[128];
jmax = 0;
while(1){
cin >> imax;
if(!imax) break;
fill(word[0], word[16], 0);
fill(digit, digit+256, 0);
fill(ok, ok+10, false);
for(int i=0; i<imax; i++) {
cin >> buffer;
k = strlen((char*)buffer) - 1;
if(k > jmax) jmax = k;
for(int j=0; j<=k; j++) {
c = word[i][j] = buffer[k - j];
if(isalpha(c)) digit[c] = -1;
else if(isdigit(c)) digit[c] = c - '0';
}
}
for(int i=0; i<=9; i++) ok[i] = true;
solution = 0;
solve(0);
cout << solution << endl;
}
return 0;
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double D;
typedef complex<D> P;
#define F first
#define S second
const ll MOD=1000000007;
//const ll MOD=998244353;
template<typename T,typename U>istream & operator >> (istream &i,pair<T,U> &A){i>>A.F>>A.S; return i;}
template<typename T>istream & operator >> (istream &i,vector<T> &A){for(auto &I:A){i>>I;} return i;}
template<typename T,typename U>ostream & operator << (ostream &o,const pair<T,U> &A){o<<A.F<<" "<<A.S; return o;}
template<typename T>ostream & operator << (ostream &o,const vector<T> &A){ll i=A.size(); for(auto &I:A){o<<I<<(--i?" ":"");} return o;}
bool solve(){
ll N;
cin>>N;
if(N==0){return false;}
vector<int> A(10);
for(int i=0;i<10;i++){A[i]=i;}
map<char,int> M;
vector<string> Str(N);
for(auto &I:Str){
cin>>I;
for(auto &J:I){M[J]=-1;}
}
//cout<<N<<endl;
//for(auto &I:Str){cout<<I<<endl;}
int cnt=0;
for(auto &I:M){I.S=cnt++;}
for(auto &I:Str){
for(auto &J:I){J=M[J];}
}
//cout<<N<<endl;
//for(auto &I:Str){for(auto &J:I){cout<<(int)J;}cout<<endl;}
ll ans=0;
do{
ll sum=0;
bool jd=true;
for(int i=0;i<N;i++){
auto &I=Str[i];
if(A[I[0]]==0 && (int)I.size()!=1){jd=false; break;}
ll k=0;
for(auto &J:I){k*=10; k+=A[J];}
sum+=(i+1==N?-1LL:1LL)*k;
}
//cout<<A<<endl;
//cout<<jd<<" "<<sum<<endl;
if(!jd || sum!=0){continue;}
ans++;
}while(next_permutation(A.begin(),A.end()));
ll uku=1;
for(ll i=10;i>cnt;i--){uku*=11-i;}
ans/=uku;
cout<<ans<<endl;
return true;
}
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
while(solve());
return 0;
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
#define LOG(...) fprintf(stderr, __VA_ARGS__)
//#define LOG(...)
#define FOR(i, a, b) for(int i = (int)(a); i < (int)(b); ++i)
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
#define RREP(i, n) for(int i = (int)(n - 1); i >= 0; --i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort(ALL(c))
#define RSORT(c) sort(RALL(c))
#define SQ(n) (n) * (n)
#define BIT(x, i) (((x) >> (i)) & 1)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef vector<ll> vll;
typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef vector<vc> vvc;
typedef vector<vll> vvll;
char c[10];
int k[10];
bool not0[10];
// vi hoge;
int m;
int dfs(int idx, int used, int lr) {
if (idx >= m) {
// if (lr == 0) {
// REP(i, hoge.size() / 3) {
// LOG("%c:%d*%d=%d ", c[i], hoge[i * 3], hoge[i * 3 + 1], hoge[i * 3 + 2]);
// }
// LOG("%d ", lr);
// LOG("\n");
// }
return lr == 0;
}
int res = 0;
FOR(i, not0[idx] ? 1 : 0, 10) {
if (BIT(used, i)) continue;
// hoge.push_back(i);
// hoge.push_back(k[idx]);
// hoge.push_back(k[idx] * i);
res += dfs(idx + 1, (used | 1 << i), lr + k[idx] * i);
// hoge.pop_back();
// hoge.pop_back();
// hoge.pop_back();
}
return res;
}
int main() {
int n;
while(cin >> n, n) {
fill_n((int*)k, 10, 0);
fill_n((bool*)not0, 10, false);
vector<string> str(n);
map<char, pair<int, bool>> alpha;
REP(i, n){
cin >> str[i];
REP(j, str[i].length()){
alpha[str[i][j]] = {0, false};
}
}
m = (int) alpha.size();
REP(i, n - 1){
for(int j = str[i].length() - 1, n = 1; j >= 0; j--, n *= 10) {
alpha[str[i][j]].first += n;
}
}
for(int j = str[str.size() - 1].length() - 1, n = 1; j >= 0; j--, n *= 10) {
alpha[str[str.size() - 1][j]].first -= n;
}
REP(i, n){
if (str[i].length() > 1)
alpha[str[i][0]].second = true;
}
int ii = 0;
for(auto it = alpha.begin(); it != alpha.end(); it++, ii++) {
c[ii] = it->first;
k[ii] = it->second.first;
not0[ii] = it->second.second;
}
// REP(i, m){
// LOG("%c %d %s\n", c[i], k[i], not0[i] ? "true" : "false");
// }
// getchar(); getchar();
cout << dfs(0, 0, 0) << endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif
//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
for (int n; cin >> n&&n;) {
vector<string> v(n);
vector<char> c;
rep(i, 0, n) {
cin >> v[i];
reverse(all(v[i]));
}
bool exist[128];
memset(exist, 0, sizeof(exist));
rep(j, 0, 20) {
rep(i, 0, n) {
if (v[i].size() <= j)continue;
if (v[i][j] >= '0'&&v[i][j] <= '9')continue;
if (exist[v[i][j]])continue;
exist[v[i][j]] = true;
c.push_back(v[i][j]);
}
}
int mp[128];
memset(mp, -1, sizeof(mp));
rep(i, 0, 10)
mp[i + '0'] = i;
bool head[128];
memset(head, 0, sizeof(head));
rep(i, 0, n) {
if (v[i].size() > 1) {
head[v[i].back()] = true;
}
}
char used[10];
memset(used, 0, sizeof(used));
function<int(int, int)> dfs = [&](int idx, int num) {
int add = 0;
rep(i, 0, 10) {
int digit = add;
add = 0;
bool flag = false;
rep(j, 0, n - 1) {
if (v[j].size() <= i)continue;
if (mp[v[j][i]] == -1) {
flag = true;
break;
}
digit += mp[v[j][i]];
}
if (flag)
break;
add += digit / 10;
digit %= 10;
if (v[n - 1].size() > i) {
if (mp[v[n - 1][i]] == -1)
break;
if (digit != mp[v[n - 1][i]])
return 0;
}
else {
if (digit > 0 || add > 0)
return 0;
else
break;
}
}
if (c.size() > idx + 1) {
int ret = 0;
rep(i, 0, 10) {
if (used[i])continue;
if (i == 0 && head[c[idx + 1]])continue;
used[i] = c[idx + 1];
mp[c[idx + 1]] = i;
ret += dfs(idx + 1, i);
mp[c[idx + 1]] = -1;
used[i] = 0;
}
return ret;
}
else {
return 1;
}
};
int ans = 0;
rep(i, 0, 10) {
if (used[i])continue;
if (i == 0 && head[c[0]])continue;
used[i] = c[0];
mp[c[0]] = i;
ans += dfs(0, i);
mp[c[0]] = -1;
used[i] = 0;
}
cout << ans << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <stdio.h>
#include <string.h>
#include <ctype.h>
using namespace std;
enum { FALSE, TRUE };
#define N 15 /* 最大の行数 */
int imax, jmax, solution, digit[256], low[256], ok[10];
char word[N][16];
void found(void){
for(int i = 0; i <= imax; i++){
if(strlen(word[i])>1&&digit[word[i][strlen(word[i])-1]]==0)return;
}
++solution;
}
void solve(int sum) /* 再帰的に試みる */
{
static int i = 0, j = 0, carry;
int c, d;
c = word[i][j];
if(i < imax) {
i++;
if((d = digit[c]) < 0) { /* 定まっていないなら */
for(d = low[c]; d <= 9; d++)
if(ok[d]) {
digit[c] = d; ok[d] = FALSE;
solve(sum + d); ok[d] = TRUE;
}
digit[c] = -1;
} else solve(sum + d);
i--;
} else {
j++; i = 0; d = sum % 10; carry = sum / 10;
if(digit[c] == d) {
if(j <= jmax) solve(carry);
else if(carry == 0) found();
} else if(digit[c] < 0 && ok[d] && d >= low[c]) {
digit[c] = d; ok[d] = FALSE;
if(j <= jmax) solve(carry);
else if(carry == 0) found();
digit[c] = -1; ok[d] = TRUE;
}
j--; i = imax;
}
}
int main(){
int i, j, k, c;
char buffer[128];
jmax = 0;
while(1){
cin >> imax;
if(!imax) break;
imax--;
memset(word, 0, sizeof(word));
memset(digit, 0, sizeof(digit));
//memset(low,0,sizeof(low));
memset(ok, 0, sizeof(ok));
for(i = 0; i < N && i <= imax; i++) {
cin >> buffer;
//low[buffer[0]] = 1;
k = strlen((char*)buffer) - 1;
if(k > jmax) jmax = k;
for(j = 0; j <= k; j++) {
c = word[i][j] = buffer[k - j];
if(isalpha(c)) digit[c] = -1;
else if(isdigit(c)) digit[c] = c - '0';
}
}
for(i = 0; i <= 9; i++) ok[i] = TRUE;
solution = 0; solve(0);
cout << solution << endl;
}
return 0;
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <complex>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
const int INF = 1<<29;
int par[10];
bool istop[10];
int m;
int num[10];
int solve() {
int res = 0;
int S = (1<<m)-1;
while(S < (1<<10)){
int p = 0;
REP(i,10)
if (S>>i&1) num[p++] = i;
do {
int sum = 0;
bool ok = 1;
REP(i,m) {
if (num[i]==0 && istop[i]) {
ok = 0;
break;
}
sum += num[i] * par[i];
}
if (ok && sum==0) res++;
}while(next_permutation(num, num+m));
int x = S & -S, y = S + x;
S = ((S & ~y) / x >> 1) | y;
}
return res;
}
int main() {
int n;
while(cin >> n , n) {
m = 0;
map<char,int> id;
memset(par,0,sizeof(par));
memset(istop,0,sizeof(istop));
REP(i, n) {
string s;
cin >> s;
int hoge = 1;
for (int j=s.size()-1; j>=0; --j) {
int index;
if (id.count(s[j])) index = id[s[j]];
else index = id[s[j]] = m++;
if (i<n-1)
par[index] += hoge;
else
par[index] -= hoge;
hoge *= 10;
}
if (s.size()>1)
istop[id[s[0]]] = 1;
}
cout << solve() << endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int n, num[26];
int A[26], B[26];
bool used[26], vis[10], fir[26];
vector<int> data;
string str[12];
int calc(int p){
int res = 0;
if(p == data.size()){
int a = 0, b = 0;
for(int i=0;i<data.size();i++){
a += num[data[i]] * A[data[i]];
b += num[data[i]] * B[data[i]];
}
if(a == b) res++;
}else{
for(int i=0;i<10;i++){
if(vis[i]) continue;
if(i == 0 && fir[data[p]]) continue;
vis[i] = true;
num[data[p]] = i;
res += calc(p+1);
vis[i] = false;
}
}
return res;
}
void pcalc(string& s, int vec[26]){
int size = s.size();
for(int i=0,r=1;i<s.size();i++,r*=10){
vec[s[size-1-i]-'A'] += r;
}
}
int main(){
while(cin >> n && n){
for(int i=0;i<26;i++){
used[i] = fir[i] = false;
A[i] = B[i] = 0;
}
for(int i=0;i<10;i++){
vis[i] = false;
}
for(int i=0;i<n;i++){
cin >> str[i];
for(int j=0;j<str[i].size();j++) used[str[i][j]-'A'] = true;
if(str[i].size() > 1) fir[str[i][0]-'A'] = true;
}
data.clear();
for(int i=0;i<26;i++) if(used[i]) data.push_back(i);
for(int i=0;i<n;i++){
if(i == n-1) pcalc(str[i], B);
else pcalc(str[i], A);
}
cout << calc(0) << endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
vector<string> in;
void init(){
in.clear();
}
int input(){
init();
int n;
cin>>n;
if(n==0)return false;
in.resize(n);
for(int i=0;i<n;i++){
cin>>in[i];
}
return true;
}
int istop[30];
void top(){
for(int i=0;i<30;i++){
istop[i]=0;
}
for(int i=0;i<in.size();i++){
if(in[i].size()!=1){
istop[in[i][0]-'A']=true;
};
}
}
bool isTop(char c){
return istop[c-'A'];
}
vector<pair<int,char>> num;
void convert(){
map<char,int> dp;
dp.clear();
for(int i=0;i<in.size();i++){
int c = 1;
if(i==in.size()-1)c*=-1;
for(int j=0;j<in[i].size();j++){
dp[in[i][in[i].size()-j-1]] += c;
c*=10;
}
}
num.clear();
for(auto i:dp){
num.push_back(make_pair(i.second,i.first));
}
sort(num.begin(),num.end());
}
vector<int> used;
int tmp;
int req(int n){
if(n==num.size()){
return tmp == 0;
}
if(n==0) tmp = 0;
if(tmp>0)return 0;
int res = 0;
for(int i=0;i<10;i++){
if(isTop(num[n].second)&&i==0)continue;
if(!used[i]){
tmp += num[n].first * i;
used[i]=true;
int tt= req(n+1);
res += tt;
used[i]=false;
tmp -= num[n].first * i;
}
}
return res;
}
int solve(){
top();
convert();
used.resize(10);
for(int i=0;i<10;i++){
used[i]=0;
}
return req(0);
}
int main(){
while(input()){
cout<<solve()<<endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | import java.util.ArrayList;
import java.util.Scanner;
public class Main {
static int n;
static String [] s;
static int[] c;
static ArrayList<Character> list=new ArrayList<>();
static boolean[] used;
static boolean[] begin;
static int dfs(int now) {
int count=0;
if(now==list.size()) {
int sum=0;
int[] sumn=new int[n];
for(int i=0; i<n-1; i++) {
for(int j=0; j<s[i].length(); j++) {
sumn[i]+=c[s[i].charAt(j)-'A'];
sumn[i]*=10;
}
sumn[i]/=10;
}
for(int i=0; i<n-1; i++) {
//System.out.println(s[i]+" "+sumn[i]);
sum+=sumn[i];
}
int ans=0;
for(int j=0; j<s[n-1].length(); j++) {
ans+=c[s[n-1].charAt(j)-'A'];
ans*=10;
}
//System.out.println(s[n-1]+" "+ans/10);
if(sum==ans/10) return 1;
else return 0;
}
for(int i=0; i<=9; i++) {
if(i==0 && begin[list.get(now)-'A']) continue;
if(used[i]) continue;
used[i]=true;
c[list.get(now)-'A']=i;
count+=dfs(now+1);
used[i]=false;
}
return count;
}
public static void main(String[] args) {
try(Scanner sc = new Scanner(System.in)){
while(sc.hasNext()) {
n=sc.nextInt();
if(n==0) break;
s=new String[n];
c=new int[27];
used=new boolean[10];
begin=new boolean[27];
boolean[] t=new boolean[27];
for(int i=0; i<n; i++) {
s[i]=sc.next();
if(s[i].length()>1) begin[s[i].charAt(0)-'A']=true;
for(int j=0; j<s[i].length(); j++) {
if(! t[s[i].charAt(j)-'A']) {
t[s[i].charAt(j)-'A']=true;
list.add(s[i].charAt(j));
}
}
}
System.out.println(dfs(0));
list.clear();
}
}
}
}
| JAVA |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <stdio.h>
#include <string.h>
int n, f[10], num[10], len[12], s[12][8], w[10], allen, c0;
char al[11];
int dfs(int c){
int i, j, k, ret = 0, t;
if(c == c0 + 1){
k = 1;
for(j = 1;j < len[n - 1];j++) k *= 10;
t = k * (num[s[n - 1][0]] + 1);
for(i = 0;i < n - 1;i++){
k = 1;
for(j = 1;j < len[i];j++) k *= 10;
t -= k * num[s[i][0]];
if(t < 0)
return 0;
}
}
if(c == allen){
int r;
t = 0;
for(k = 0;k < len[n - 1];k++)
t = t * 10 + num[s[n - 1][k]];
r = t;
for(j = 0;j < n - 1;j++){
t = 0;
for(k = 0;k < len[j];k++)
t = t * 10 + num[s[j][k]];
r -= t;
if(r < 0) break;
}
return r==0?1:0;
}
for(i = 0;i < 10;i++){
if(!(w[c] && i == 0) && !f[i]){
f[i] = 1;
num[c] = i;
ret += dfs(c + 1);
f[i] = 0;
}
}
return ret;
}
int main(void){
int i, j, k, cnt, m;
char c, str[12][9];
while(1){
scanf("%d%*c",&n);
if(!n) break;
for(i = 0;i < n;i++){
j = 0;
while((c = getchar()) != '\n')
str[i][j++] = c;
str[i][j] = '\0';
}
for(i = 0;i < n;i++)
len[i] = strlen(str[i]);
allen = 0;
m = 0;
do{
for(j = 0;j < n;j++){
if(len[j] > m){
for(k = 0;k < allen;k++)
if(str[j][m] == al[k]) break;
if(k == allen){
al[k] = str[j][m], al[k + 1] = '\0';
allen++;
}
s[j][m] = k;
}
}
cnt = 0;
for(i = 0;i < n;i++)
if(len[i] > m) cnt++;
m++;
}while(cnt);
memset(w,0,sizeof(w));
for(i = 0;i < n;i++)
if(len[i] > 1) w[s[i][0]] = 1;
c0 = 0;
for(i = 0;i < n;i++)
if(s[i][0] > c0) c0 = s[i][0];
memset(f,0,sizeof(f));
printf("%d\n",dfs(0));
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <cstring>
#include <vector>
#include <sstream>
#include <map>
using namespace std;
#define index foo
map<char, int> index;
int ch_count;
int coef[11];
int as[11];
bool not_zero[11];
int pow10(int n) {
int ret = 1;
for (int i = 0; i < n; i++)
ret *= 10;
return ret;
}
bool check() {
int sum = 0;
for (int i = 1; i < ch_count; i++) {
sum += as[i] * coef[i];
}
//if (sum == 0) {
// for (int i = 1; i < ch_count; i++) {
// cout << as[i] << ' ' << coef[i] << endl;
// }
// cout << endl;
//}
return sum == 0;
}
int cnt(int cur = 1, int used = 0) {
if (cur == ch_count) return check() ? 1 : 0;
int ret = 0;
for (int i = 0; i < 10; i++) {
if (i == 0 && not_zero[cur]) continue;
if (used & (1 << i)) continue;
as[cur] = i;
ret += cnt(cur+1, used | (1 << i));
as[cur] = 0;
}
return ret;
}
int main() {
int n;
while (cin >> n, n) {
int cur = 1;
index = map<char, int>();
memset(coef, 0, sizeof(coef));
memset(as, 0, sizeof(as));
memset(not_zero, 0, sizeof(not_zero));
for (int i = 0; i < n-1; i++) {
string e; cin >> e;
for (int j = 0; j < e.size(); j++) {
if (index[e[j]] == 0) index[e[j]] = cur++;
coef[index[e[j]]] += pow10(e.size()-1-j);
}
if (e.size() != 1) not_zero[index[e[0]]] = true;
}
string obj;
cin >> obj;
for (int j = 0; j < obj.size(); j++) {
if (index[obj[j]] == 0) index[obj[j]] = cur++;
coef[index[obj[j]]] -= pow10(obj.size()-1-j);
}
if (obj.size() != 1) not_zero[index[obj[0]]] = true;
ch_count = cur;
cout << cnt() << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | import java.util.*;
public class Main{
int n, k, cnt;
String[] string;
ArrayList<Character> head, appear;
int[] num;
long[] coeff;
void solve(){
Scanner sc = new Scanner(System.in);
while(true){
n = sc.nextInt();
if(n==0) break;
string = new String[n];
for(int i=0; i<n; i++) string[i] = sc.next();
head = new ArrayList<Character>();
appear = new ArrayList<Character>();
for(int i=0; i<n; i++){
char ch = string[i].charAt(0);
if(!head.contains(ch) && string[i].length()!=1) head.add(ch);
for(int j=0; j<string[i].length(); j++){
char c = string[i].charAt(j);
if(!appear.contains(c)) appear.add(c);
}
}
k = appear.size();
coeff = new long[k];
long coe = 1;
for(int i=0; i<n; i++){
coe = 1;
for(int j=string[i].length()-1; j>=0; j--){
if(i!=n-1)coeff[appear.indexOf(string[i].charAt(j))] += coe;
else coeff[appear.indexOf(string[i].charAt(j))] -= coe;
coe *= 10;
}
}
cnt = 0;
num = new int[k];
dfs(0,new int[10],0);
System.out.println(cnt);
}
}
void dfs(int pos, int[] used, long sumL){
if(pos==k){
if(equal()) cnt++;
return;
}
long sumR = 0;
if(sumL>0){
for(int i=pos; i<k; i++) if(coeff[i]<0) sumR += coeff[i]*(-9);
if(sumL>sumR) return;
}else{
for(int i=pos; i<k; i++) if(coeff[i]>0) sumR += coeff[i]*9;
if(-sumL>sumR) return;
}
for(int i=0; i<10; i++){
if(used[i]==0){
if(i==0 && head.contains(appear.get(pos))) continue;
num[pos] = i;
used[i] = 1;
dfs(pos+1,used,sumL+coeff[pos]*num[pos]);
used[i] = 0;
}
}
}
boolean equal(){
long sum = 0;
for(int i=0; i<k; i++) sum += coeff[i]*num[i];
return sum==0 ? true:false;
}
public static void main(String[] args){
new Main().solve();
}
} | JAVA |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
bool used[10];
bool tabu[10];
int keisu[10];
int cnt, ans;
void dfs(int k, int sum){
if (k == cnt) {
if (!sum) {
ans++;
}
return;
}
for (int i = 0; i < 10; i++) {
if (used[i] || (i == 0 && tabu[k])) {
continue;
}
used[i] = true;
dfs(k+1, sum+i*keisu[k]);
used[i] = false;
}
return;
}
int main(){
while (true) {
int n;
cin >> n;
if (n == 0) {
break;
}
vector<string> str(n);
map<char, int> id;
cnt = 0;
ans = 0;
for (int i = 0; i < 10; i++) {
used[i] = false;
tabu[i] = false;
keisu[i] = 0;
}
for(int i = 0; i < n; i++){
cin >> str[i];
string s = str[i];
for (int j = 0; j < s.size(); j++) {
if (id.find(s[j]) == id.end()) {
id[s[j]] = cnt++;
}
if (j == 0 && s.size() > 1) { // 0にならない記号か
tabu[id[s[j]]] = true;
}
if (i != n - 1) {
keisu[id[s[j]]] += (int)pow(10, s.size() - j - 1);
}else{
keisu[id[s[j]]] -= (int)pow(10, s.size() - j - 1);
}
}
}
dfs(0, 0);
cout << ans << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
template<class T> inline T sqr(T x) {return x*x;}
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define each(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define exist(s,e) ((s).find(e)!=(s).end())
#define range(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n) range(i,0,n)
#define clr(a,b) memset((a), (b) ,sizeof(a))
#define dump(x) cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
const double eps = 1e-10;
const double pi = acos(-1.0);
const ll INF =1LL << 62;
const int inf =1 << 29;
ll coef[26];
int from[26];
string in[15];
inline int cut(int i,int j){
int cur=i;
rep(loop,j) cur/=10;
return cur%10;
}
int main(void){
int n;
while(cin >> n,n){
int ans=0;
rep(i,n) cin >> in[i];
clr(coef,0);
clr(from,-1);
rep(i,n)rep(j,in[i].size()){
int m=in[i].size();
int cur=in[i][j]-'A';
from[cur]=max(from[cur],(m>1&&j==0)?1:0);
if(i<n-1)
coef[cur]+=pow(10,m-1-j);
else
coef[cur]-=pow(10,m-1-j);
}
vi index;
rep(i,26) if(from[i]!=-1) index.pb(i);
int num=index.size();
rep(mask,1<<10){
if(__builtin_popcount(mask)!=num) continue;
vi number;
rep(j,10) if(mask&(1<<j)) number.pb(j);
do{
bool check=true;
rep(j,num) if(number[j]<from[index[j]]) check=false;
if(!check) continue;
int csum=0;
rep(j,num) csum+=number[j]*coef[index[j]];
if(csum==0) ans++;
}while(next_permutation(number.begin(),number.end()));
}
cout << ans << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
string str[13];
int used[10],num[101],n,flg[101];
int nxp(int idx,int i,int sum,int a){
int res=0;
if(sum<0||idx==n)return sum==0;
if(i<0)return nxp(idx+1,str[idx+1].size()-1,sum,1);
int f=idx? -1:1,ch=str[idx][i];
if(num[ch]!=-1) res+=nxp(idx,i-1,sum+f*(a*num[ch]),a*10);
else
for(int j=0;j<10;j++){
if(used[j]||(!i&&!j&&str[idx].size()!=1)||(!j&&flg[ch]))continue;
used[j]=1,num[ch]=j;
res+=nxp(idx,i-1,sum+f*(a*j),a*10);
used[j]=0,num[ch]=-1;
}
return res;
}
int main(){
memset(num,-1,sizeof(num));
while(cin>>n,n){
memset(flg,0,sizeof(flg));
for(int i=0;i<n;i++){
cin>>str[i];
if(str[i].size()!=1) flg[str[i][0]]=1;
}
swap(str[0],str[n-1]);
cout <<nxp(0,str[0].size()-1,0,1)<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<char> VC;
typedef vector<VC> VVC;
#define Y first
#define X second
#define RANGE(x, y, mX, mY) (0 <= (x) && 0 <= (y) && (x) < (mX) && (y) < (mY))
#define MP make_pair
const int INF = 0x3f3f3f3f;
int N, M;
vector<string> ss;
set<char> not_zero;
VC as;
map<char, int> ks;
// alphabet?????????
VC find_alphas() {
set<char> as_set;
for (auto s : ss) {
for (auto c : s) {
as_set.insert(c);
}
}
VC res;
for (auto c : as_set) {
res.push_back(c);
}
return res;
}
void find_not_zero() {
not_zero.clear();
for (auto &s : ss) {
if (s.size() == 1) continue;
not_zero.insert(s[0]);
}
}
// ?????°?¨????
void calc_ks() {
ks.clear();
for (auto c : as) {
ks[c] = 0;
}
for (int i = 0; i < N; i++) {
int v = 1;
string &s = ss[i];
for (int j = (int)s.size() - 1; j >= 0; --j) {
ks[s[j]] += v * (i == N - 1 ? -1 : 1);
v *= 10;
}
}
}
bool calc_siki(VI &idxes) {
int sum = 0;
for (int i = 0; i < M; i++) {
char c = as[i];
int n = idxes[i];
sum += ks[c] * n;
}
return sum == 0;
}
int dfs(int used, int n, int sum) {
if (n == M) {
return sum == 0;
}
int cnt = 0;
for (int i = 0; i < 10; i++) {
if (1 << i & used) continue;
if (!i && not_zero.count(as[n])) continue;
cnt += dfs(used | 1 << i, n + 1, sum + ks[as[n]] * i);
}
return cnt;
}
int solve() {
as = find_alphas();
M = as.size();
find_not_zero();
calc_ks();
return dfs(0, 0, 0);
}
int main(void) {
cin.tie( 0 );
ios::sync_with_stdio( false );
while (cin >> N, N) {
ss.clear();
ss.resize(N);
for (auto &s : ss) cin >> s;
cout << solve() << endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #define _USE_MATH_DEFINES
#define INF 100000000
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>
using namespace std;
typedef long long ll;
typedef pair <string,string> P;
typedef pair <int,P> PP;
static const double EPS = 1e-8;
void dfs(map<char,int>& co,map<char,int>::iterator it,map<char,bool>& pre,int visited,int sum,map<int, vector<int> >& result){
if(it==co.end()){
result[sum].push_back(visited);
return;
}
for(int i=0;i<=9;i++){
if(visited & (1<<i)) continue;
if(i==0 && pre[it->first]) continue;
sum += it->second * i;
it++;
visited |= (1<<i);
dfs(co,it,pre,visited,sum,result);
visited &= ~(1<<i);
it--;
sum -= it->second * i;
}
}
int main(){
int n;
while(~scanf("%d",&n)){
if(n==0) break;
int count = 0;
map<char,int> coefficient;
map<char,bool> prefix; // prefix or not
for(int i=0;i<n;i++){
string str;
cin >> str;
if(str.size() > 1) prefix[str[0]] = true;
for(int j=str.size()-1,k=1;j>=0;j--,k*=10){
if(i!=n-1) coefficient[str[j]] += k;
else coefficient[str[j]] -= k;
}
}
map<char,int> co_front;
map<char,int> co_rear;
int ub=0;
map<char,int>::iterator cit;
for(cit = coefficient.begin(); cit != coefficient.end();cit++){
if(ub >= coefficient.size()/2) break;
co_front[cit->first] = cit->second;
ub++;
}
for(; cit != coefficient.end();cit++){
co_rear[cit->first] = cit->second;
}
map<int,vector<int> > result_front;
map<int,vector<int> > result_rear;
dfs(co_front,co_front.begin(),prefix,0,0,result_front);
dfs(co_rear,co_rear.begin(),prefix,0,0,result_rear);
int res=0;
for(map<int,vector<int> >::iterator it = result_front.begin(); it != result_front.end(); it++){
map<int,vector<int> >::iterator it2;
if((it2 = result_rear.find(-(it->first))) != result_rear.end()){
for(vector<int>::iterator sit = it->second.begin(); sit != it->second.end(); sit++){
for(vector<int>::iterator sit2 = it2->second.begin(); sit2 != it2->second.end(); sit2++){
if(!(*sit & *sit2)) res++;
}
}
}
}
printf("%d\n",res);
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 |
// http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1161&lang=jp
#include<iostream>
#include<map>
#include<vector>
#include<algorithm>
#include<cmath>
#include<climits>
#include<ctime>
#include<cstring>
#include<numeric>
#define ALL(v) (v).begin(),(v).end()
#define REP(i,p,n) for(int i=p;i<(int)(n);++i)
#define rep(i,n) REP(i,0,n)
#define dump(a) (cerr << #a << "=" << (a) << endl)
#define DUMP(list) cout << "{ "; for(auto nth : list){ cout << nth << " "; } cout << "}" << endl;
using namespace std;
vector<int> coefficient;
vector<bool> is_top;
int dfs(int m, int depth, int used, int value) {
if(depth == m)
return ((value == 0) ? 1 : 0);
int ret = 0;
rep(i, 10) {
if(is_top[depth] && i == 0) continue;
if(used & (1 << i)) continue;
ret += dfs(m, depth + 1, used | (1 << i), value + coefficient[depth] * i);
}
return ret;
}
int main() {
int n;
while(cin >> n, n) {
map<char, int> converter;
// アルファベットの種類を数える
vector<string> s(n, "");
rep(i, n) {
cin >> s[i];
rep(j, s[i].size()) {
if(!converter.count(s[i][j])) converter.insert(make_pair(s[i][j], converter.size()));
}
}
// 各アルファベットの係数と先頭にあるかを求める
int m = converter.size();
coefficient.assign(m, 0);
is_top.assign(m, false);
rep(i, n) {
for(int j = s[i].size() - 1, k = 1; j >= 0; --j, k *= 10) {
int ctoi = converter[s[i][j]];
if(j == 0 && s[i].size() > 1) is_top[ctoi] = true;
coefficient[ctoi] += (i == n - 1) ? -k : k;
}
}
// 覆面算が成り立つ式を満たす組み合わせを探索
cout << dfs(m, 0, 0, 0) << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
bool isfirst[26];
bool isused[26];
int N;
while(cin>>N,N){
fill(isfirst,isfirst+26,false);
fill(isused,isused+26,false);
vector<string> S;
int K[26]={};
for(int i=0;i<N;i++){
string str;
cin>>str;
S.push_back(str);
if(str.size()>1) isfirst[str[0]-'A']=true;
for(auto c:str) isused[c-'A']=true;
}
for(int n=0;n<N-1;n++){
for(int i=0,k=1;i<8;i++,k*=10){
K[S[n][S[n].size()-1-i]-'A']+=k;
}
}
for(int i=0,k=1;i<8;i++,k*=10){
K[S[N-1][S[N-1].size()-1-i]-'A']-=k;
}
int used=0;
for(int i=0;i<26;i++) if(isused[i])used++;
int ans=0;
int comb=(1<<used)-1;
while(comb<(1<<10)){
int perm[10];
for(int i=0,p=0;i<10;i++){
if(comb&(1<<i)){
perm[p++]=i;
}
}
int x = comb&-comb, y=comb+x;
comb=((comb&~y)/x>>1)|y;
do{
int num[26];
int d,sum=0;
for(int i=0,p=0;i<26;i++){
if(isused[i]){
num[i]=perm[p++];
if(isfirst[i]&&num[i]==0){
sum=-1;
break;
}
sum+=K[i]*num[i];
}
}
ans+=!sum;
}while(next_permutation(perm,perm+used));
}
cout<<ans<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using PII = pair<int, int>;
#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back
signed main(void)
{
cin.tie(0);
ios::sync_with_stdio(false);
while(true) {
int n;
cin >> n;
if(!n) break;
vector<string> s(n);
REP(i, n) cin >> s[i];
// 座圧しておく
vector<char> v;
REP(i, n) REP(j, s[i].size()) v.PB(s[i][j]);
sort(ALL(v)); v.erase(unique(ALL(v)), v.end());
REP(i, n) REP(j, s[i].size()) s[i][j] = lower_bound(ALL(v), s[i][j]) - v.begin() + 'A';
// leading-zero対策
vector<bool> fro(n, false);
REP(i, n) if(s[i].size() >= 2) fro[s[i][0]-'A'] = true;
// 係数を求める
VI a(15, 0);
REP(i, n) {
for(int j=s[i].size()-1, tmp=1; j>=0; --j, tmp*=10) {
if(i == n-1) a[s[i][j]-'A'] -= tmp;
else a[s[i][j]-'A'] += tmp;
}
}
map<PII, int> mp1, mp2;
int m = v.size(), m2 = m/2;
function<void(int,int,int)> dfs1 = [&](int x, int sum, int used) {
if(x == m2) {mp1[{sum, used}]++; return;}
REP(i, 10) {
// leading-zeroとかすでに使ってる数は使えない
if(fro[x] && i==0) continue;
if(used&1<<i) continue;
dfs1(x+1, sum+i*a[x], used | 1<<i);
}
};
function<void(int,int,int)> dfs2 = [&](int x, int sum, int used) {
if(x == m) {mp2[{sum, used}]++; return;}
REP(i, 10) {
if(fro[x] && i==0) continue;
if(used&1<<i) continue;
dfs2(x+1, sum+i*a[x], used | 1<<i);
}
};
// 列挙
dfs1(0, 0, 0);
dfs2(m2, 0, 0);
vector<pair<PII, int>> vec1(ALL(mp1)), vec2(ALL(mp2));
for(auto &i: vec2) i.first.first *= -1;
// 尺取りをする
int ans = 0;
int l = 0, r = vec2.size()-1;
for(;l<vec1.size();) {
while(r > 0 && vec1[l].first.first > vec2[r].first.first) r--;
int val1 = vec1[l].first.first, val2 = vec2[r].first.first;
// 合計が0にならないならcontinue
if(val1 != val2) {
while(l < vec1.size() && vec1[l].first.first==val1) l++;
continue;
}
// 合計が0になるような組み合わせを全て見て使ってる数が被っていなければ答えにプラス
for(int i=l; i<vec1.size() && vec1[i].first.first==val1; ++i) {
for(int j=r; j>=0 && vec2[j].first.first==val2; --j) {
if((vec1[i].first.second&vec2[j].first.second) == 0) {
ans += vec1[i].second * vec2[j].second;
}
}
}
while(l < vec1.size() && vec1[l].first.first==val1) l++;
}
cout << ans << endl;
}
return 0;
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define mp(a,b) make_pair(a,b)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> PPI;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;
#define INF INT_MAX/3
int n;
string s[13];
map<char,int> alc,dame;
bool used[13];
int res;
void dfs(int i,int d,int sum){
if(i==n-1&&d==0){
if(alc[s[i][d]]==sum||(alc[s[i][d]]==-1&&sum<10&&sum!=0&&!used[sum])){
res++;
}
}else if(i==n-1){
if(alc[s[i][d]]==sum%10){
dfs(0,d-1,sum/10);
}else if(alc[s[i][d]]==-1&&!used[sum%10]&&(sum%10!=0||!dame[s[i][d]])){
used[sum%10]=true;
alc[s[i][d]]=sum%10;
dfs(0,d-1,sum/10);
used[sum%10]=false;
alc[s[i][d]]=-1;
}
}else if(alc[s[i][d]]==-1){
repl(j,dame[s[i][d]],10){
if(used[j])continue;
alc[s[i][d]]=j;
used[j]=true;
dfs(i+1,d,sum+j);
alc[s[i][d]]=-1;
used[j]=false;
}
}else{
dfs(i+1,d,sum+alc[s[i][d]]);
}
return ;
}
int main(){
cin.sync_with_stdio(false);
while(1){
cin>>n;
if(n==0)break;
alc.clear(); dame.clear();
int maxl=0;
rep(i,n){
cin>>s[i];
rep(j,s[i].length()){
if(j==0&&s[i].length()!=1)dame[s[i][j]]=1;
alc[s[i][j]]=-1;
}
maxch(maxl,(int)s[i].length());
}
rep(i,n){
while(s[i].length()<maxl)s[i]="-"+s[i];
}
memset(used,0,sizeof(used));
alc['-']=0;
res=0;
dfs(0,maxl-1,0);
cout<<res<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | /*
* 1161.cc: Verbal Arithmetic
*/
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
using namespace std;
/* constant */
const int MAX_N = 12;
const int MAX_M = 8;
/* typedef */
/* global variables */
int n, m, gn;
int ids[26], ds[10], exps[MAX_N][MAX_M];
bool zfs[10], used[10];
/* subroutines */
int rec(int x, int y, int co) {
if (x >= m) return (co == 0) ? 1 : 0;
int eid = exps[y][x];
if (y == n - 1) {
int sum = co;
for (int i = 0; i < n - 1; i++)
sum += (exps[i][x] >= 0) ? ds[exps[i][x]] : 0;
int dn = sum % 10, de = ds[eid];
if (de < 0) {
if (! used[dn] && (! zfs[eid] || dn > 0)) {
used[dn] = true;
ds[eid] = dn;
int r = rec(x + 1, 0, sum / 10);
ds[eid] = -1;
used[dn] = false;
return r;
}
}
else if (dn == de) return rec(x + 1, 0, sum / 10);
return 0;
}
if (eid < 0 || ds[eid] >= 0) return rec(x, y + 1, co);
int r = 0;
for (int d = 0; d <= 9; d++)
if (! used[d] && (! zfs[eid] || d > 0)) {
used[d] = true;
ds[eid] = d;
r += rec(x, y + 1, co);
ds[eid] = -1;
used[d] = false;
}
return r;
}
/* main */
int main() {
for (;;) {
cin >> n;
if (n == 0) break;
memset(ids, -1, sizeof(ids));
memset(ds, -1, sizeof(ds));
memset(zfs, false, sizeof(zfs));
memset(used, false, sizeof(used));
gn = m = 0;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
int sn = s.size();
if (m < sn) m = sn;
for (int j = 0; j < sn; j++) {
int ch = s[sn - 1 - j] - 'A';
if (ids[ch] < 0) ids[ch] = gn++;
exps[i][j] = ids[ch];
}
for (int j = sn; j < MAX_M; j++) exps[i][j] = -1;
if (sn > 1) zfs[exps[i][sn - 1]] = true;
}
if (false) {
for (int i = 0; i < n; i++) {
for (int j = m - 1; j >= 0; j--) {
if (exps[i][j] >= 0) printf("%d", exps[i][j]);
else putchar(' ');
}
putchar('\n');
}
for (int i = 0; i < gn; i++) printf("%d", zfs[i]); putchar('\n');
}
int ans = rec(0, 0, 0);
printf("%d\n", ans);
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #define _USE_MATH_DEFINES
#include <iostream>
#include <complex>
#include <algorithm>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <cmath>
#include <math.h>
#include <numeric>
#include <list>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <climits>
#include <set>
#include <memory.h>
#include <memory>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cassert>
#include <map>
#include <cassert>
#include <time.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<int , P> PP;
typedef pair<int, PP> PPP;
typedef vector<int> vec;
typedef vector<vec> mat;
typedef complex<double> Point;
const int INF= 1 << 30;
const double EPS = 1e-9;
const double PI = 3.1415926535897932384626433832795;
bool operator < (const Point & a, const Point & b){
return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
}
int res;
bool use[10];
int s1[10];
int s2[10];
int num[10];
map<char, int> mp;
map<char, bool> used;
map<int , bool> bad;
void dfs(int n, int val){
if(n == 0){
if(val == 0){
res++;
}
return ;
}
if(val > 0 && val + s1[n - 1] > 0) return;
if(val <= 0 && val + s2[n - 1] < 0) return ;
for(int i = 0; i < 10; i++){
if(i == 0 && bad[n - 1]) continue;
if(!use[i]){
use[i] = true;
dfs(n - 1, val + i * num[n - 1]);
use[i] = false;
}
}
}
int pow(int n){
int res = 1;
for(int i = 0; i < n; i++) res *= 10;
return res;
}
int main(){
int n;
while(cin >> n && n){
string s;
int c = 0;
fill(num, num + 10, 0);
mp.clear();
bad.clear();
used.clear();
for(int i = 0; i < n; i++){
cin >> s;
for(int j = 0; j < (int)s.size(); j++){
if(!used[s[s.size() - 1 - j]]){
mp[s[s.size() - 1 - j]] = c++;
used[s[s.size() - 1 - j]] = true;
}
ll d = pow(j);
if(i == n - 1){
num[mp[s[s.size() - 1 - j]]] -= d;
}else{
num[mp[s[s.size() - 1 - j]]] += d;
}
if(j == s.size() - 1 && s.size() > 1) bad[mp[s[0]]] = true;
}
}
s1[0] = min(0, num[0]) * 9;
s2[0] = max(0, num[0]) * 9;
for(int i = 1; i < c; i++){
s1[i] = s1[i-1] + min(0, num[i]) * 9;
s2[i] = s2[i-1] + max(0, num[i]) * 9;
}
res = 0;
memset(use, false, sizeof(use));
dfs(c, 0);
cout << res << endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
const ll MOD = 1000000007;
const int INF = 1<<29;
int N;
vector<string> strs;
set<int> non_zero;
int mapped_num[26];
bool used[10];
int max_len;
int dfs(int x, int y, int s){
if(x >= max_len){
if(s == 0) return 1;
else return 0;
}
if(y == N - 1){
if(x >= strs[y].size()) return 0;
int d = strs[y][strs[y].size()-1-x] - 'A';
if(mapped_num[d] == -1){
int i = s % 10;
if(used[i]) return 0;
if(i == 0 && non_zero.count(d) > 0) return 0;
used[i] = true;
mapped_num[d] = i;
int ret = dfs(x+1, 0, s / 10);
mapped_num[d] = -1;
used[i] = false;
return ret;
}else{
if(mapped_num[d] != s % 10) return 0;
return dfs(x+1, 0, s / 10);
}
}else{
if(x >= strs[y].size()) return dfs(x, y+1, s);
int d = strs[y][strs[y].size()-1-x] - 'A';
if(mapped_num[d] != -1) return dfs(x, y+1, s+mapped_num[d]);
int ret = 0;
for(int i=0;i<10;i++){
if(used[i]) continue;
if(i == 0 && non_zero.count(d) > 0) continue;
used[i] = true;
mapped_num[d] = i;
ret += dfs(x, y+1, s+i);
mapped_num[d] = -1;
used[i] = false;
}
return ret;
}
}
int main(){
while(true){
cin >> N;
if(N == 0) break;
strs.resize(N);
non_zero.clear();
max_len = 0;
for(int i=0;i<N;i++){
cin >> strs[i];
if(strs[i].size() > 1) non_zero.insert(strs[i][0] - 'A');
max_len = max(max_len, int(strs[i].size()));
}
memset(mapped_num, -1, sizeof(mapped_num));
memset(used, 0, sizeof(used));
cout << dfs(0, 0, 0) << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | import java.awt.geom.Point2D;
import java.io.IOException;
import java.io.InputStream;
import java.util.*;
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.function.Supplier;
public class Main {
final static int INF = 1 << 28;
final static long MOD = 1_000_000_007;
final static double EPS = 1e-9;
final static double GOLDEN_RATIO = (1.0 + Math.sqrt(5)) / 2.0;
Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
new Main().run();
}
int n, m;
ArrayList<String> list;
boolean[] done;
char[] check;
int[] map;
int[] list2;
HashMap<Character, ArrayList<Node>> map2;
HashSet<Character> top;
int sum1;
class Node {
int index0;
int index1;
}
int decode(String str) {
int time = 1;
int sum = 0;
for (int i = str.length() - 1; 0 <= i; --i) {
char c = str.charAt(i);
sum += map[c] * time;
time *= 10;
}
return sum;
}
boolean ok() {
// boolean flag = true;
// for (int i = 0; i < n; ++i) {
// if (list.get(i).length() > 1) {
// flag &= map[list.get(i).charAt(0)] != 0;
// }
// }
// int sum1 = 0;
// for (int i = 0; i < n - 1; ++i) {
//// sum1 += decode(list.get(i));
// sum1 += list2[i];
// }
// int sum2 = decode(list.get(n - 1));
int sum2 = list2[n - 1];
// return sum1 == sum2 && flag;
return sum1 == sum2;
}
int dfs(int depth) {
if (depth == m) {
return ok() ? 1 : 0;
}
int cnt = 0;
for (int i = 0; i < 10; ++i) {
if (done[i]) {
continue;
}
if (i == 0 && top.contains(check[depth])) {
continue;
}
done[i] = true;
map[check[depth]] = i;
for (Node node : map2.get(check[depth])) {
list2[node.index0] += i * node.index1;
if (node.index0 < n - 1) {
sum1 += i * node.index1;
}
}
cnt += dfs(depth + 1);
for (Node node : map2.get(check[depth])) {
list2[node.index0] -= i * node.index1;
if (node.index0 < n - 1) {
sum1 -= i * node.index1;
}
}
done[i] = false;
}
return cnt;
}
void run() {
map = new int[256];
done = new boolean[10];
for (; ; ) {
n = ni();
if (n == 0) {
break;
}
list = new ArrayList<>();
HashSet<Character> set = new HashSet<>();
map2 = new HashMap<>();
top = new HashSet<>();
for (int i = 0; i < n; ++i) {
String str = sc.next();
list.add(str);
int j = 1;
for (int k = 0; k < str.length(); ++k) {
j *= 10;
}
for (char c : str.toCharArray()) {
set.add(c);
Node node = new Node();
node.index0 = i;
node.index1 = j;
if (!map2.containsKey(c)) {
map2.put(c, new ArrayList<>());
}
map2.get(c).add(node);
j /= 10;
}
if (str.length() > 1) {
top.add(str.charAt(0));
}
}
ArrayList<Character> check = new ArrayList<>(set);
m = check.size();
this.check = new char[m];
for (int i = 0; i < m; ++i) {
this.check[i] = check.get(i);
}
list2 = new int[n];
sum1 = 0;
int ans = dfs(0);
System.out.println(ans);
}
}
int ni() {
return Integer.parseInt(sc.next());
}
void debug(Object... os) {
System.err.println(Arrays.deepToString(os));
}
/**
* ??????????????????????????????
*
* @return a ??¨ b ????????§??¬?´???°
*/
long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
/**
* ????????????????????????????????????
*
* @return mx + ny = gcd(m, n)??¨???????????????(x, y)?????????
*/
Pair<Long, Long> gcd_ex(long m, long n) {
long[][] mat = _gcd_ex(m, n);
return new Pair<>(mat[0][0], mat[0][1]);
}
long[][] _gcd_ex(long m, long n) {
if (n == 0) {
return new long[][]{{1, 0}, {0, 1}};
}
long k = m / n;
long[][] K = new long[][]{{0, 1}, {1, -k}};
long[][] r = _gcd_ex(n, m % n);
long[][] dst = new long[2][2];
for (int y = 0; y < 2; ++y)
for (int x = 0; x < 2; ++x)
for (int i = 0; i < 2; ++i)
dst[y][x] += r[y][i] * K[i][x];
return dst;
}
/**
* ??°?????????2???????????¨??????????????????????£?
*
* @return a^r (mod 1,000,000,007)
*/
long pow(long a, long r) {
long sum = 1;
while (r > 0) {
if ((r & 1) == 1) {
sum *= a;
sum %= MOD;
}
a *= a;
a %= MOD;
r >>= 1;
}
return sum;
}
/**
* ???????????????
* O(n)
*
* @return {}_nC_r
*/
long C(int n, int r) {
long sum = 1;
for (int i = n; 0 < i; --i) {
sum *= i;
sum %= MOD;
}
long s = 1;
for (int i = r; 0 < i; --i) {
s *= i;
s %= MOD;
}
sum *= pow(s, MOD - 2);
sum %= MOD;
long t = 1;
for (int i = n - r; 0 < i; --i) {
t *= i;
t %= MOD;
}
sum *= pow(t, MOD - 2);
sum %= MOD;
return sum;
}
/**
* ??????????????¢?´¢
*
* @param left ??????
* @param right ??????
* @param f ??¢?´¢????????¢??°
* @param comp ??????????????¢??°?????¢?´¢????????¨?????????Comparator.comparingDouble(Double::doubleValue)
* ??????????????¢??°?????¢?´¢????????¨?????????Comparator.comparingDouble(Double::doubleValue).reversed()
* @return ?\?????????§?¨?x
*/
double goldenSectionSearch(double left, double right, Function<Double, Double> f, Comparator<Double> comp) {
double c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
double c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
double d1 = f.apply(c1);
double d2 = f.apply(c2);
while (right - left > 1e-9) {
if (comp.compare(d1, d2) > 0) {
right = c2;
c2 = c1;
d2 = d1;
c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
d1 = f.apply(c1);
} else {
left = c1;
c1 = c2;
d1 = d2;
c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
d2 = f.apply(c2);
}
}
return right;
}
/**
* [a,b]???m:n???????????????????????????
*/
double divideInternally(double a, double b, double m, double n) {
return (n * a + m * b) / (m + n);
}
/**
* http://alexbowe.com/popcount-permutations/
* bit????????£???????????°????°?????????????????????????????????¨???????????????
* ex)
* <pre>
* for (int i = 0; i < 25; ++i) {
* int bits = (1 << i) - 1;
* long m = C(25, num);
* for (j = 0; j < m; ++j) {
* ...(25???????????????i???bit????????£?????????)
* if (bits != 0)
* bits = next_perm(bits);
* }
* }
* </pre>
*
* @param v ?????¨???bit???
* @return ?¬????bit???
*/
int next_perm(int v) {
int t = (v | (v - 1)) + 1;
return t | ((((t & -t) / (v & -v)) >> 1) - 1);
}
/**
* from http://gihyo.jp/dev/serial/01/geometry part 6
*/
static class Line {
double a;
double b;
double c;
/**
* ?????¬??¢??????????????????????????´??????????????????
*
* @param a x????????°
* @param b y????????°
* @param c ?????°???
*/
Line(double a, double b, double c) {
this.a = a;
this.b = b;
this.c = c;
}
/**
* 2???(x1, y1), (x2, y2)???????????´??????????????????
*
* @param x1 1?????????x??§?¨?
* @param y1 1?????????y??§?¨?
* @param x2 2?????????x??§?¨?
* @param y2 2?????????y??§?¨?
* @return ??´???
*/
static Line fromPoints(double x1, double y1, double x2, double y2) {
double dx = x2 - x1;
double dy = y2 - y1;
return new Line(dy, -dx, dx * y1 - dy * x1);
}
/**
* ?????????????????´?????¨??????????????????
*
* @param l ??´???
* @return ?????????2??´?????????????????´??????null
*/
Point2D getIntersectionPoint(Line l) {
double d = a * l.b - l.a * b;
if (d == 0.0) {
return null;
}
double x = (b * l.c - l.b * c) / d;
double y = (l.a * c - a * l.c) / d;
return new Point2D.Double(x, y);
}
@Override
public String toString() {
return "a = " + a + ", b = " + b + ", c = " + c;
}
}
/**
* from http://gihyo.jp/dev/serial/01/geometry part 6
*/
static public class LineSegment {
double x1;
double y1;
double x2;
double y2;
LineSegment(double x1, double y1, double x2, double y2) {
this.x1 = x1;
this.y1 = y1;
this.x2 = x2;
this.y2 = y2;
}
Line toLine() {
return Line.fromPoints(x1, y1, x2, y2);
}
boolean intersects(Line l) {
double t1 = l.a * x1 + l.b * y1 + l.c;
double t2 = l.a * x2 + l.b * y2 + l.c;
return t1 * t2 <= 0;
}
boolean intersects(LineSegment s) {
return bothSides(s) && s.bothSides(this);
}
// s???????????????????????´????????????????????????????????????
private boolean bothSides(LineSegment s) {
double ccw1 = GeomUtils.ccw(x1, y1, s.x1, s.y1, x2, y2);
double ccw2 = GeomUtils.ccw(x1, y1, s.x2, s.y2, x2, y2);
if (ccw1 == 0 && ccw2 == 0) { // s??¨?????????????????´?????????????????´???
// s???????????????1???????????????????????????????????????????????°???s??????????????¨??±?????¨?????????????????§
// true?????????
return internal(s.x1, s.y1) || internal(s.x2, s.y2);
} else { // ????????\????????´???
// CCW???????¬?????????°????????´??????true?????????
return ccw1 * ccw2 <= 0;
}
}
// (x, y)?????????????????????????????????????????????????????????
private boolean internal(double x, double y) {
// (x, y)????????????????????????????????????????????????????????\?????§????????°????????¨?????????
return GeomUtils.dot(x1 - x, y1 - y, x2 - x, y2 - y) <= 0;
}
public Point2D getIntersectionPoint(Line l) {
if (!intersects(l)) {
return null; // ?????????????????´??????null?????????
}
return l.getIntersectionPoint(toLine());
}
public Point2D getIntersectionPoint(LineSegment s) {
if (!intersects(s)) {
return null; // ?????????????????´??????null?????????
}
return s.toLine().getIntersectionPoint(toLine());
}
/**
* from : http://www.deqnotes.net/acmicpc/2d_geometry/lines#distance_between_line_segment_and_point
*/
double distance(double x0, double y0) {
// ???????????§??????
if (GeomUtils.dot(x2 - x1, y2 - y1, x0 - x1, y0 - y1) < EPS) {
return GeomUtils.abs(x0 - x1, y0 - y1);
}
if (GeomUtils.dot(x1 - x2, y1 - y2, x0 - x2, y0 - y2) < EPS) {
return GeomUtils.abs(x0 - x2, y0 - y2);
}
// ??´?????¨???????????¢
return Math.abs(GeomUtils.cross(x2 - x1, y2 - y1, x0 - x1, y0 - y1)) / GeomUtils.abs(x2 - x1, y2 - y1);
}
double distance(LineSegment l) {
if (this.intersects(l)) {
return 0.0;
}
double min = Double.MAX_VALUE;
min = Math.min(min, distance(l.x1, l.y1));
min = Math.min(min, distance(l.x2, l.y2));
min = Math.min(min, l.distance(x1, y1));
min = Math.min(min, l.distance(x2, y2));
return min;
}
@Override
public String toString() {
return "(" + x1 + ", " + y1 + ") - (" + x2 + ", " + y2 + ")";
}
}
/**
* from http://gihyo.jp/dev/serial/01/geometry part 6
*/
static class GeomUtils {
static double cross(double x1, double y1, double x2, double y2) {
return x1 * y2 - x2 * y1;
}
static double dot(double x1, double y1, double x2, double y2) {
return x1 * x2 + y1 * y2;
}
// (x1, y1) -> (x2, y2) -> (x3, y3) ??¨?????????????????????????¨????????????´????????£?????????
// ????¨????????????´???????????????????????´???????????´?????????????????????
static double ccw(double x1, double y1, double x2, double y2,
double x3, double y3) {
return cross(x2 - x1, y2 - y1, x3 - x2, y3 - y2);
}
static double ccw(Point2D p1, Point2D p2, Point2D p3) {
return ccw(p1.getX(), p1.getY(), p2.getX(), p2.getY(), p3.getX(), p3.getY());
}
static double abs(double x, double y) {
return Math.sqrt(x * x + y * y);
}
}
/**
* http://qiita.com/p_shiki37/items/65c18f88f4d24b2c528b
*/
static class FastScanner {
private final InputStream in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
public FastScanner(InputStream in) {
this.in = in;
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
} else {
ptr = 0;
try {
buflen = in.read(buffer);
} catch (IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte()) return buffer[ptr++];
else return -1;
}
private void skipUnprintable() {
while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
}
public boolean hasNext() {
skipUnprintable();
return hasNextByte();
}
public String next() {
if (!hasNext()) throw new NoSuchElementException();
StringBuilder sb = new StringBuilder();
int b = readByte();
while (isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext()) throw new NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
n *= 10;
n += b - '0';
} else if (b == -1 || !isPrintableChar(b)) {
return minus ? -n : n;
} else {
throw new NumberFormatException();
}
b = readByte();
}
}
}
static class Pair<F extends Comparable<F>, S extends Comparable<S>> implements Comparable<Pair<F, S>> {
F f;
S s;
Pair() {
}
Pair(F f, S s) {
this.f = f;
this.s = s;
}
Pair(Pair<F, S> p) {
f = p.f;
s = p.s;
}
@Override
public int compareTo(Pair<F, S> p) {
if (f.compareTo(p.f) != 0) {
return f.compareTo(p.f);
}
return s.compareTo(p.s);
}
@Override
public int hashCode() {
return f.hashCode() ^ s.hashCode();
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || this.f == null || this.s == null) {
return false;
}
if (this.getClass() != o.getClass()) {
return false;
}
Pair p = (Pair) o;
return this.f.equals(p.f) && this.s.equals(p.s);
}
@Override
public String toString() {
return "{" + f.toString() + ", " + s.toString() + "}";
}
}
class BIT<T> {
int n;
ArrayList<T> bit;
BiFunction<T, T, T> bif;
/**
* 1-indexed ???Binary Indexed Tree????§??????????
*
* @param n ??????
* @param bif ?????¨???????????¢??°
* @param sup ?????????
*/
BIT(int n, BiFunction<T, T, T> bif, Supplier<T> sup) {
this.n = n;
bit = new ArrayList<>(n + 1);
for (int i = 0; i < n + 1; ++i) {
bit.add(sup.get());
}
this.bif = bif;
}
/**
* i??????????????????v??§??´??°??????
*
* @param i index
* @param v ??°?????????
*/
void set(int i, T v) {
for (int x = i; x <= n; x += x & -x) {
bit.set(x, bif.apply(bit.get(x), v));
}
}
/**
* ?????¨??????
*
* @param defaultValue ?????????
* @param i index
* @return [1, i]?????§f????????¨????????????
*/
T reduce(T defaultValue, int i) {
T ret = defaultValue;
for (int x = i; x > 0; x -= x & -x) {
ret = bif.apply(ret, bit.get(x));
}
return ret;
}
}
class SegmentTree<T> {
int n;
ArrayList<T> dat;
BiFunction<T, T, T> bif;
Supplier<T> sup;
/**
* 0-indexed ???Segment Tree????§??????????
*
* @param n_ ????±???????
* @param bif ?????¨???????????¢??°
* @param sup ?????????
*/
SegmentTree(int n_, BiFunction<T, T, T> bif, Supplier<T> sup) {
n = 1;
while (n < n_) n *= 2;
dat = new ArrayList<>(2 * n - 1);
for (int i = 0; i < 2 * n - 1; ++i) {
dat.add(sup.get());
}
this.bif = bif;
this.sup = sup;
}
/**
* k??????????????????v??§??´??°??????
*
* @param k index
* @param v ??°?????????
*/
void set(int k, T v) {
k += n - 1;
dat.set(k, v);
while (k > 0) {
k = (k - 1) / 2;
dat.set(k, bif.apply(dat.get(k * 2 + 1), dat.get(k * 2 + 2)));
}
}
/**
* ?????¨??????
*
* @param l ?????????
* @param r ?????????
* @return [l, r)??§?????????bif????????¨?????????????????????
*/
T reduce(int l, int r) {
return _reduce(l, r, 0, 0, n);
}
T _reduce(int a, int b, int k, int l, int r) {
if (r <= a || b <= l) return sup.get();
if (a <= l && r <= b) return dat.get(k);
T vl = _reduce(a, b, k * 2 + 1, l, (l + r) / 2);
T vr = _reduce(a, b, k * 2 + 2, (l + r) / 2, r);
return bif.apply(vl, vr);
}
}
class UnionFind {
int[] par;
UnionFind(int n) {
par = new int[n];
for (int i = 0; i < n; ++i) {
par[i] = i;
}
}
int find(int x) {
if (par[x] == x) {
return x;
}
return par[x] = find(x);
}
boolean same(int x, int y) {
return find(x) == find(y);
}
void union(int x, int y) {
x = find(x);
y = find(y);
if (x == y) {
return;
}
par[x] = y;
}
}
} | JAVA |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>
using namespace std;
#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
int main(void) {
int n;
while(cin >> n, n){
vs s(n);
rep(i,n) cin >> s[i];
int co[26] = {};
int c[26] = {};//leading zeros
rep(i,n){
if(s[i].size()>1)c[s[i][0]-'A']++;
int t[26] = {};
rep(j,s[i].size()){
rep(k,26)t[k]*=10;
t[s[i][j]-'A']++;
}
if(i+1<n)rep(j,26)co[j] += t[j];
else rep(j,26)co[j] -= t[j];
}
int a[26] = {};
rep(i,n)rep(j,s[i].size())a[s[i][j] - 'A']++;
vi v;
rep(i,26)if(a[i])v.push_back(i);
while(v.size()<10)v.push_back(-1);
sort(all(v));
int ans = 0;
do{
int alph[26];
rep(i,26) alph[i] = -1;
rep(i,v.size())if(v[i] >= 0)alph[v[i]]=i;
if(v[0] >= 0 && c[v[0]])continue;
int sum = 0;
rep(i,26)if(alph[i]>=0){
sum += co[i] * alph[i];
}
if(sum == 0){
ans++;
}
}while(next_permutation(all(v)));
cout << ans << endl;
}
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,n1,n2) for(int i=n1;i<n2;i++)
#define bFOR(i,n1,n2) for(int i=n1;i>=n2;i--)
#define speed_up ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
typedef long long int ll;
typedef pair<ll,ll> Pi;
const int INF=(ll)(1LL<<30)-1;
const double INFd=100000000000.0;
const ll INFl=(ll)9223372036854775807;
const int MAX=10000;
const ll MOD=(ll)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a, b)*b;}
int dx[4]={0,-1,0,1},dy[4]={-1,0,1,0};
int ddx[6]={0,1,0,-1,1,-1},ddy[6]={-1,0,1,0,1,-1};
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
std::fill( (T*)array, (T*)(array+N), val );
}
int n;
string s[12];
int ca[10];//alphabet-num
char c[10];//alphabet
int nc;// how many alphabet
int sk[10];//firts number or not
int nsk;//how many ↑
int plu[10];
int w[8]={1,10,100,1000,10000,100000,1000000,10000000};
int re_ans=0;
int nm_alr[10]={};
void dfs(int p){
if(p==nc){
int ans=0;
REP(i,nc){
ans+=plu[i]*ca[i];
}
if(ans==0){
re_ans++;
}
return;
}
for(int i=0;i<10;i++){
if(nm_alr[i]==1)continue;
if(i==0&&sk[p])continue;
ca[p]=i;
nm_alr[i]=1;
//cout<<"3"<<endl;
dfs(p+1);
nm_alr[i]=0;
}
return;
}
vector<int> rere_ans;
int main(){
while(1){
cin>>n;
if(n==0)break;
nc=0;
re_ans=0;
nsk=0;
nc=0;
REP(i,10)nm_alr[i]=0;
REP(i,n)
cin>>s[i];
REP(i,n){
REP(j,s[i].size()){
for(int k=0;k<nc;k++){
if(c[k]==s[i][j])goto A;
}
c[nc++]=s[i][j];
A:;
}
}
//cout<<nc<<endl;
REP(i,10)plu[i]=0;
REP(i,n-1){
REP(j,s[i].size()){
int tt=s[i].size()-j-1;
for(int k=0;k<nc;k++){
if(c[k]==s[i][j]){
plu[k]+=w[tt];
break;
}
}
}
}
REP(j,s[n-1].size()){
int tt=s[n-1].size()-j-1;
for(int k=0;k<nc;k++){
if(c[k]==s[n-1][j]){
plu[k]=plu[k]-w[tt];
break;
}
}
}
REP(i,10)sk[i]=0;
REP(i,n){
if(s[i].size()==1)continue;
REP(j,nc){
if(c[j]==s[i][0])sk[j]=1;
}
}
//REP(i,nc)cout<<c[i]<<" "<<plu[i]<<endl;
dfs(0);
rere_ans.push_back(re_ans);
}
REP(i,rere_ans.size())cout<<rere_ans[i]<<endl;
return 0;
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<iostream>
#include<iomanip>
#include<math.h>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<unordered_map>
#define REP(i, N) for(ll i = 0; i < N; ++i)
#define FOR(i, a, b) for(ll i = a; i < b; ++i)
#define ALL(a) (a).begin(),(a).end()
#define pb push_back
#define INF (long long)1000000000
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int main(void) {
while(true) {
int N;
cin>>N;
if(N == 0) break;
int size = 0;
ll res = 0;
vector<string> s(N);
map<char, ll> omomil;
map<char, ll> omomir;
int perm[10];
vector<int> out;
REP(i, 10) {
perm[i] = (int)i;
}
REP(i, N) cin>>s[i];
REP(i, N) {
REP(j, s[i].size()) {
if(omomil.count(s[i][s[i].size() - 1 - j]) == 0) {
omomil[s[i][s[i].size() - 1 - j]] = 0;
omomir[s[i][s[i].size() - 1 - j]] = 0;
++size;
}
if(i != N - 1) omomil[s[i][s[i].size() - 1 - j]] += pow(10, j);
else omomir[s[i][s[i].size() - 1 - j]] += pow(10, j);
}
}
map<char, ll>::iterator ite3 = omomil.begin();
int foo = 0;
while(ite3 != omomil.end()) {
REP(i, N) {
if(s[i].size() > 1 && s[i][0] == (*ite3).first) {
out.pb(foo);
//cout<<(*ite3).first<<":"<<foo<<endl;
break;
}
}
++ite3;
++foo;
}
ll sugoil[10];
ll sugoir[10];
ite3 = omomil.begin();
REP(i, size) {
sugoil[i] = (*ite3).second;
++ite3;
}
ite3 = omomir.begin();
REP(i, size) {
sugoir[i] = (*ite3).second;
++ite3;
}
do {
ll l = 0;
ll r = 0;
bool cont = false;
int roc = 0;
REP(i, out.size()) {
if(perm[out[i]] == 0) {
cont = true;
break;
}
}
if(cont) continue;
while(roc < size) {
l += perm[roc] * sugoil[roc];
++roc;
}
roc = 0;
while(roc < size) {
r += perm[roc] * sugoir[roc];
++roc;
}
if(l == r) ++res;
} while(next_permutation(perm, perm + 10));
ll hoge = 1;
REP(i, 10 - size) {
hoge *= 10 - size - i;
}
cout<<res / hoge<<endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | //=================================
// Created on: 2018/07/04 15:12:12
//=================================
#include <bits/stdc++.h>
using namespace std;
int main()
{
for (int N;;) {
cin >> N;
if (N == 0) { break; }
vector<int> coeff1(10, 0);
vector<int> coeff2(10, 0);
set<int> nz;
map<char, int> ind;
for (int i = 0; i < N; i++) {
string s;
cin >> s;
reverse(s.begin(), s.end());
for (int j = 0, d = 1; j < s.size(); j++, d *= 10) {
const char c = s[j];
if (ind.find(c) == ind.end()) {
const int s = ind.size();
ind[c] = s;
}
const int in = ind[c];
(i < N - 1 ? coeff1 : coeff2).at(in) += d;
if (j == s.size() - 1 and s.size() > 1) { nz.insert(in); }
}
}
assert(ind.size() <= 10);
vector<int> num{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int div = 1;
for (int i = 0; i < 10 - (int)ind.size(); i++) { div *= (i + 1); }
int ans = 0;
do {
bool ok = true;
for (const int nzz : nz) {
if (num[nzz] == 0) {
ok = false;
break;
}
}
if (not ok) { continue; }
int left = 0, right = 0;
for (int i = 0; i < ind.size(); i++) { left += coeff1[i] * num[i], right += coeff2[i] * num[i]; }
if (left == right) { ans++; }
} while (next_permutation(num.begin(), num.end()));
cout << ans / div << endl;
}
return 0;
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<iostream>
#include<cstring>
#include<cstdlib>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
typedef long long ll;
string id;
ll sum[22], ans;
bool head[22], use[22];
void rec(int depth, ll val){
if(depth+1 == (int)id.size()){
ll rem = sum[depth];
if(rem == 0LL){
if(val == 0LL){
ans += 10-depth;
if(!use[0] && head[depth])ans--;
}
}else{
if(val == 0LL){
if(!use[0] && !head[depth])ans++;
}else if(val*rem<0){
val = abs(val); rem = abs(rem);
if(val%rem==0LL && val/rem<10LL && !use[val/rem]){
ans++;
}
}
}
return;
}
for(ll i=0;i<10;i++){
if(i==0LL && head[depth])continue;
if(use[i])continue;
use[i] = true;
rec(depth+1, val + sum[depth]*i);
use[i] = false;
}
}
int main(){
int n;
string s;
while(cin >> n, n){
memset(sum,0,sizeof(sum));
memset(head,0,sizeof(head));
id = "";
rep(i,n){
cin >> s;
ll tmp = 1;
for(int j=(int)s.size()-1;j>=0;j--){
int k=0;
for(;k<(int)id.size();k++){
if(s[j] == id[k])break;
}
sum[k] += (i+1==n)?-tmp:tmp;
if(k==id.size())id += s[j];
if(j==0 && s.size()!=1)head[k] = true;
tmp *= 10LL;
}
}
ans = 0;
memset(use,0,sizeof(use));
rec(0,0);
cout << ans << endl;
}
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int n;
int value[128];
vector<string> strs;
int ans = 0;
bool leading[128];
bool used[10];
int to_num(const string &s) {
int res = 0;
for(const auto &c : s) {
res = res * 10 + value[c];
}
return res;
}
void dfs(int keta,int tate,int sum){
if(keta == -1){
ans++;
/* for(const auto &s : strs) {
cout << to_num(s) << endl;
}
cout << endl;
*/
return ;
}
char m = strs[tate][keta];
if(tate == n-1){
if(value[m] == -1){
if(sum%10 == 0 && leading[m] || used[sum%10])return ;
value[m] = sum%10;
used[sum%10] = true;
dfs(keta-1,0,sum/10);
used[sum%10] = false;
value[m] = -1;
}
else if(value[m] == sum%10){
dfs(keta-1,0,sum/10);
}
return ;
}
if(value[m] != -1){
dfs(keta,tate+1,sum+value[m]);
}else{
for(int i = leading[m]? 1:0;i < 10;i++){
if(used[i])continue;
used[i] = true;
value[m] = i;
dfs(keta,tate+1,sum+value[m]);
used[i] = false;
}
value[m] = -1;
}
}
int main(){
while(cin >> n && n){
fill(begin(value),end(value),-1);
fill(begin(leading),end(leading),false);
fill(begin(used),end(used),false);
ans = 0;
strs.clear();
value['#'] = 0;
for(int i = 0;i < n;i++){
string tmp;
cin >> tmp;
if(tmp.size() > 1){
leading[tmp[0]] = true;
}
strs.emplace_back((string(8,'#')+tmp).substr(tmp.size()));
}
dfs(7,0,0);
cout << ans << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string P[20];
int N, c[26], F[20], r[20]; char d[26]; int cnt1 = 0;
int X10[3628800];
int fact(int n) {
int sum1 = 1;
for (int i = 1; i <= n; i++)sum1 *= i;
return sum1;
}
int main() {
int a2[10] = { 0,1,2,3,4,5,6,7,8,9 }, cnt4 = 0;
do {
for (int i = 0; i < 10; i++) {
if (!a2[i]) X10[cnt4] = i;
}
cnt4++;
} while (next_permutation(a2, a2 + 10));
while (true) {
for (int i = 0; i <= 10; i++) { F[i] = fact(i); r[i] = 0; }
for (int i = 0; i < 20; i++) { P[i] = ""; }
for (int i = 0; i < 26; i++) { c[i] = 0; d[i] = ' '; }
cnt1 = 0;
cin >> N; if (N == 0) { break; }
for (int i = 0; i < N; i++) {
cin >> P[i];
for (int j = 0; j < P[i].size(); j++) {
c[P[i][j] - 'A']++;
}
}
for (int i = 0; i < 26; i++) {
if (c[i] >= 1) {
d[i] = cnt1 + 'A';
cnt1++;
}
else {
d[i] = 'Z';
}
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < P[i].size(); j++) {
P[i][j] = d[P[i][j] - 'A'];
}
if (P[i].size() >= 2) { r[P[i][0] - 'A'] = 1; }
}
int a[10] = { 0,1,2,3,4,5,6,7,8,9 };
int cnt2 = 0, cnt3 = 0, V = 0, sum = 0;
int R[100];
for (int i = 0; i < N; i++)R[i] = P[i].size();
do {
if (cnt2 % F[10 - cnt1]) { cnt2++; continue; }
if (r[X10[cnt2]] == 1) { cnt2++; continue; }
sum = 0;
for (int i = 0; i < N; i++) {
int Q = 0;
for (int j = 0; j < R[i]; j++) Q = Q * 10 + a[P[i][j] - 'A'];
if (i != N - 1) sum += Q;
else if (sum == Q) cnt3++;
}
E:;
cnt2++;
}
while (next_permutation(a, a + 10));
cout << cnt3 << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
#define int long long
vector<int> z,s;
vector<string> vs;
int rec(int p,int u,int sum){
if(p==z.size()) return !sum;
int res=0;
for(int i=z[p];i<10;i++) if(!((u>>i)&1)) res+=rec(p+1,u|(1<<i),sum+s[p]*i);
return res;
}
signed main(){
int n,i,j,k;
while(cin>>n,n){
vs.clear();vs.resize(n);
for(i=0;i<n;i++) cin>>vs[i];
vector<char> vc;
map<char,int> m;
for(i=0;i<vs.size();i++)
for(j=0;j<vs[i].size();j++)
if(m.find(vs[i][j])==m.end())
m[vs[i][j]]=vc.size(),vc.push_back(vs[i][j]);
z.clear();z.resize(vc.size(),0);
s.clear();s.resize(vc.size(),0);
reverse(vs.begin(),vs.end());
for(i=0;i<vs.size();i++){
if(vs[i].size()-1) z[m[vs[i][0]]]=1;
reverse(vs[i].begin(),vs[i].end());
for(j=0,k=1;j<vs[i].size();j++) s[m[vs[i][j]]]+=(i?1:-1)*k,k*=10;
}
cout << rec(0,0,0) << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> table(26, -1);
vector<bool> used(10, false);
int dfs(int d, vector<string> &str, vector<char> &clist, vector<bool> &nonzero, vector<long long int> &num){
if(d==0){
long long int sum=0;
for(int i=0; i<(int)clist.size(); i++){
sum += num[i]*table[clist[i]-'A'];
}
if(sum==0){
return 1;
}else{
return 0;
}
}
int ret = 0;
for(int i=0; i<=9; i++){
if(used[i]) continue;
if(i==0 && nonzero[clist[d-1]-'A']) continue;
used[i] = true;
table[clist[d-1]-'A'] = i;
ret += dfs(d-1, str, clist, nonzero, num);
table[clist[d-1]-'A'] = -1;
used[i] = false;
}
return ret;
}
int main(){
while(1){
int n;
cin >> n;
if(n==0) break;
vector<string> str(n);
vector<char> clist;
vector<bool> nonzero(26, false);
for(int i=0; i<n; i++){
cin >> str[i];
for(int j=0; j<(int)str[i].length(); j++){
clist.push_back(str[i][j]);
}
if((int)str[i].length()!=1){
nonzero[str[i][0]-'A'] = true;
}
}
sort(clist.begin(), clist.end());
clist.erase(unique(clist.begin(), clist.end()), clist.end());
int numc = clist.size();
vector<long long int> num(numc, 0);
for(int d=0; d<numc; d++){
for(int i=0; i<n; i++){
long long int sub = 0;
for(int j=0; j<(int)str[i].length(); j++){
sub *= 10;
if(str[i][j] == clist[d]) sub+=1;
}
if(i==n-1) num[d] -= sub;
else num[d] += sub;
}
}
cout << dfs(numc, str, clist, nonzero, num) << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
// #include <iostream>
// #include <set>
// #include <map>
// #include <vector>
using namespace std;
int n;
string s[15];
int main() {
// cin.tie(0);
// ios::sync_with_stdio(false);
while(true) {
cin>>n;
int ans=0;
int chr[26]={};
bool isused[26]={};
int coef_l[26]={},coef_r[26]={};
bool istop[26]={};
if(!n) break;
for(int i=0; i<n; ++i) {
cin>>s[i];
if(s[i].size()>1) istop[s[i][0]-'A']=true;
reverse(s[i].begin(),s[i].end());
long long j=1;
for(char c:s[i]) {
isused[c-'A']=true;
if(i<n-1) coef_l[c-'A']+=j;
else coef_r[c-'A']+=j;
j*=10;
}
}
int k=0;
for(int i=0; i<26; ++i) {
if(isused[i]) ++k;
}
int p[10]={0,1,2,3,4,5,6,7,8,9};
int lasp[10]={-1};
do {
bool suc=false;
for(int i=0; i<k; ++i) {
if(lasp[i]!=p[i]) {
lasp[i]=p[i];
suc=true;
}
}
if(!suc) continue;
long long le=0,ri=0;
for(int i=0,j=0; i<26; ++i) {
if(isused[i]) {
if(p[j]==0 && istop[i]) suc=false;
le+=coef_l[i]*p[j];
ri+=coef_r[i]*p[j];
++j;
}
}
if(suc && (le==ri)) {
ans++;
}
} while(next_permutation(p,p+10));
cout<<ans<<'\n';
}
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#define int long long
using namespace std;
class Solver {
};
vector<vector<int>> permutation(vector<int> factors, vector<bool> used, int cnt) {
vector<vector<int>> ret;
if (cnt == 1) {
for (int i = 0; i < factors.size(); i++) {
if (used[i])continue;
vector<int> tmp({ factors[i] });
ret.emplace_back(tmp);
}
return ret;
}
for (int i = 0; i < factors.size(); i++) {
if (used[i])continue;
used[i] = true;
vector<vector<int>> tmp = permutation(factors, used, cnt - 1);
for (int j = 0; j < tmp.size(); j++) {
tmp[j].emplace_back(factors[i]);
ret.emplace_back(tmp[j]);
}
used[i] = false;
}
return ret;
}
signed main() {
while (true) {
int n;
cin >> n;
if (n == 0)break;
vector<string> strs(n);
for (int i = 0; i < n; i++) {
cin >> strs[i];
}
map<char, int> indexes;
set<char> heads;
for (int i = 0; i < n; i++) {
int div = 1;
for (int j = strs[i].size() - 1; j >= 0; j--) {
if (indexes.find(strs[i][j]) == indexes.end()) {
indexes.emplace(strs[i][j], 0);
}
if (i == n - 1)indexes[strs[i][j]] -= div;
else indexes[strs[i][j]] += div;
div *= 10;
if (j == 0 && strs[i].size() > 1)heads.emplace(strs[i][j]);
}
}
vector<int> factors(10);
for (int i = 0; i < 10; i++)factors[i] = i;
//vector<bool> used(10, false);
//vector<vector<int>> pats = permutation(factors, used, indexes.size());
int res = 0;
do {
int val = 0;
int j = 0;
bool valid = true;
for (auto key : indexes) {
if (factors[j] == 0 && heads.find(key.first) != heads.end()) {
valid = false;
break;
}
val += key.second * factors[j];
j++;
}
if (valid && val == 0)res++;
} while (next_permutation(factors.begin(), factors.end()));
int divider = 1;
for (int i = 1; i <= 10 - indexes.size(); i++)divider *= i;
cout << res/divider << endl;
}
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
bool D[10],used[10];
int kei[10],m,an;
void dfs(int x,int sum){
if(x==m){
if(sum==0)an++;
return;
}
for(int i=0;i<10;i++){
if((i==0&&D[x])||used[i])continue;
used[i]=1;
dfs(x+1,sum+kei[x]*i);
used[i]=0;
}
}
int main(){
int n;
while(cin>>n&&n){
m=0,an=0;
memset(D,0,sizeof(D));
string s;
map<char,int> c;
vector<char> v;
for(int i=0;i<n;i++){
cin>>s;
int l=1;
if(i==n-1)l=-1;
for(int j=s.size()-1,k=1;j>=0;j--,k*=10){
if(j==0&&s.size()!=1)v.push_back(s[j]);
if(c.find(s[j])==c.end())c[s[j]]=l*k;
else c[s[j]]+=l*k;
}
}
map<char,int>::iterator it;
for(it=c.begin();it!=c.end();it++){
for(int i=0;i<v.size();i++)
if((*it).first==v[i])D[m]=1;
kei[m++]=(*it).second;
}
dfs(0,0);
cout<<an<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<iostream>
#include<algorithm>
using namespace std;
int num[200];
bool flg[200];
int n,m,ans;
int t[20][20];
bool used[10];
void solve(int y,int x,int sum){
if(y==0&&x==m){
if(sum==0)ans++;
return;
}
int a=num[t[y][x]];
if(y==n-1){
if(a==-1){
if((!flg[t[y][x]]||sum%10!=0) && !used[sum%10]){
used[sum%10]=true;
num[t[y][x]]=sum%10;
solve(0,x+1,sum/10);
used[sum%10]=false;num[t[y][x]]=-1;
}
}else if(a==sum%10)solve(0,x+1,sum/10);
}else{
if(a==-1){
for(int i=0;i<10;i++){
if(i==0&&flg[t[y][x]])continue;
if(used[i])continue;
num[t[y][x]]=i;
used[i]=true;
solve(y+1,x,sum+i);
num[t[y][x]]=-1;used[i]=false;
}
}else solve(y+1,x,sum+a);
}
}
void init(){
m=ans=0;
for(int i=0;i<200;i++)num[i]=-1,flg[i]=false;
for(int i=0;i<20;i++)for(int j=0;j<20;j++)t[i][j]=0;
for(int i=0;i<10;i++)used[i]=false;
num[0]=0;
}
int main(){
while(1){
cin>>n;
if(n==0)break;
init();
string str;
for(int i=0;i<n;i++){
cin>>str;
int size=str.size();
m=max(m,size);
for(int j=0;j<size;j++)t[i][j]=str[size-j-1];
if(size>1)flg[(int)str[0]]=true;
}
solve(0,0,0);
cout<<ans<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<numeric>
#include<map>
#include<unordered_map>
#include<set>
#include<stack>
#include<queue>
#define reps(i,j,k) for(int i=(j);i<(k);i++)
#define rep(i,j) reps(i,0,j)
#define fs first
#define sc second
#define pb push_back
#define mk make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
template<class S,class T>
ostream &operator<<(ostream &out, const pair<S,T> p){
return out << "(" << p.fs << ", " << p.sc << ")";
}
template<class T>
ostream &operator<<(ostream &out, const vector<T> &v){
out << "{";
rep(i,v.size()) out << v[i] << ", ";
return out << "}" << endl;
}
int main(){
int n;
while(cin >> n, n){
vector<string> v(n);
vector<char> c;
vi num(10);
rep(i,10) num[i] = i;
rep(i,n){
cin >> v[i];
rep(j,v[i].length()){
char ch = v[i][j];
rep(k,c.size()) if(ch == c[k]){
ch = 0;
break;
}
if(ch) c.pb(ch);
}
}
while(c.size() < 10) c.pb(0);
sort(c.begin(), c.end());
int cnt = 0;
vi istop(256);
rep(i,n){
if(v[i].length() == 1) continue;
istop[v[i][0]] = 1;
}
vi keisu(256);
rep(i,n){
int t = 1;
reverse(v[i].begin(), v[i].end());
rep(j, v[i].length()){
keisu[v[i][j]] += i != n-1 ?t : -t;
t *= 10;
}
}
do{
if(istop[c[0]]) continue;
int sum = 0;
rep(i,10) sum += keisu[c[i]] * i;
cnt += sum == 0;
}while(next_permutation(c.begin(), c.end()));
cout << cnt << endl;;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<utility>
#include<numeric>
#include<algorithm>
#include<bitset>
#include<complex>
#include<stack>
using namespace std;
typedef long long Int;
typedef vector<int> vint;
typedef pair<int,int> pint;
#define mp make_pair
template<class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cout << *i << " "; cout << endl; }
template<class T> void chmin(T &t, T f) { if (t > f) t = f; }
template<class T> void chmax(T &t, T f) { if (t < f) t = f; }
int in() { int x; scanf("%d", &x); return x; }
#define rep(i,n) for(int i=0;i<(n);i++)
int main() {
int n;
string str[20];
int alpha[30];
int al[30];
int top[30];
for(;;){
memset(alpha,0,sizeof(alpha));
memset(al,0,sizeof(alpha));
memset(top,0,sizeof(alpha));
n=in();
if(n==0)break;
rep(i,n)cin>>str[i];
rep(i,n){
int ten=1;
for(int j=str[i].size()-1;j>=0;j--,ten*=10){
al[str[i][j]-'A']=1;
if(i<n-1)alpha[str[i][j]-'A']+=ten;
else alpha[str[i][j]-'A']-=ten;
}
if(str[i].size()>1)top[str[i][0]-'A']=1;
}
int used[30];
int m=0;
rep(i,30){
if(al[i]>0)used[m++]=i;
}
int res=0;
do{
int bit=(1<<m)-1;
int q;
for(;bit<1<<10;q=bit+(bit&-bit),bit=q|((bit&~q)/(bit&-bit))>>1){
if((bit&1)>0&&top[used[0]]>0)continue;
int c=0,sum=0;
rep(i,10){
if((bit&1<<i)!=0){
sum+=alpha[used[c++]]*i;
}
}
if(sum==0)res++;
}
}while(next_permutation(used,used+m));
cout<<res<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
int main(){
while (1){
int N;
cin >> N;
if (N == 0){
break;
}
vector<string> S(N);
for (int i = 0; i < N; i++){
cin >> S[i];
}
vector<char> ch;
for (int i = 0; i < N; i++){
int sz = S[i].size();
for (int j = 0; j < sz; j++){
ch.push_back(S[i][j]);
}
}
sort(ch.begin(), ch.end());
ch.erase(unique(ch.begin(), ch.end()), ch.end());
int cnt = ch.size();
map<char, char> mp;
for (int i = 0; i < cnt; i++){
mp[ch[i]] = 'A' + i;
}
for (int i = 0; i < N; i++){
int sz = S[i].size();
for (int j = 0; j < sz; j++){
S[i][j] = mp[S[i][j]];
}
}
vector<bool> ok(cnt, true);
vector<int> sum(cnt, 0);
for (int i = 0; i < N; i++){
int sz = S[i].size();
if (sz > 1){
ok[S[i][0] - 'A'] = false;
}
int c = 1;
for (int j = sz - 1; j >= 0; j--){
if (i < N - 1){
sum[S[i][j] - 'A'] += c;
} else {
sum[S[i][j] - 'A'] -= c;
}
c *= 10;
}
}
vector<int> p(10);
for (int i = 0; i < 10; i++){
p[i] = i;
}
int ans = 0;
while (1){
int d = 0;
bool ok2 = true;
for (int j = 0; j < cnt; j++){
d += sum[j] * p[j];
if (p[j] == 0 && !ok[j]){
ok2 = false;
}
}
if (d == 0 && ok2){
ans++;
}
if (!next_permutation(p.begin(), p.end())){
break;
}
}
for (int i = 1; i <= 10 - cnt; i++){
ans /= i;
}
cout << ans << endl;
}
}
| CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
int n,num[101],used[10],size,flg[101];
string str[13];
int B[10];
int nxp(int idx,int i,int sum){
if(sum<0||i==size) return sum==0;
if(idx==n) return nxp(0,i+1,sum);
int f=idx? -1:1,ch=str[idx][i],res=0;
if(num[ch]!=-1) return nxp(idx+1,i,sum+f*(B[size-i-1]*num[ch]));
for(int j=0;j<10;j++){
if(used[j]||(!j&&flg[ch]))continue;
used[j]=1,num[ch]=j;
res+=nxp(idx+1,i,sum+f*(B[size-i-1]*j));
used[j]=0,num[ch]=-1;
}
return res;
}
int solve(){
str[0]='0'+str[0];
size=str[0].size();
for(int i=1;i<n;i++){
if(size<str[i].size()) return 0;
while(size>str[i].size()) str[i]='0'+str[i];
}
return nxp(0,0,0);
}
int main(){
B[0]=1;
for(int i=1;i<10;i++)B[i]=B[i-1]*10;
memset(num,-1,sizeof(num));
num['0']=0;
while(cin>>n,n){
memset(flg,0,sizeof(flg));
for(int i=0;i<n;i++){
cin>>str[i];
if(str[i].size()>1)flg[str[i][0]]=1;
}
swap(str[0],str[n-1]);
cout <<solve()<<endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
#define all(v) (v).begin(), (v).end()
vector<char> alpha;
int coef[28];
bool nonzero[28], used[10];
int solve(int idx, int sum)
{
if(idx == alpha.size()) return sum == 0;
int ret = 0, c = alpha[idx] - 'A';
for(int i = 0; i < 10; i++) {
if(used[i]) continue;
if(i == 0 && nonzero[c]) continue;
used[i] = true;
ret += solve(idx + 1, sum + coef[c]*i);
used[i] = false;
}
return ret;
}
int main()
{
int N;
while(cin >> N, N) {
alpha.clear();
fill(coef, coef + 28, 0);
fill(nonzero, nonzero + 28, false);
fill(used, used + 10, false);
for(int i = 0; i < N; i++) {
string item; cin >> item;
if(item.size() != 1) nonzero[item[0]-'A'] = true;
reverse(all(item));
int mul = 1;
for(char a : item) {
alpha.push_back(a);
coef[a-'A'] += (i == N-1 ? -mul : mul);
mul *= 10;
}
}
sort(all(alpha));
alpha.erase(unique(all(alpha)), alpha.end());
cout << solve(0, 0) << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int fact(int n){
if (n == 0)
return 1;
else
return n * fact(n - 1);
}
int main()
{
int len[13];
int * np[150];
char alphabet[150];
int n;
int num[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
//入力
while (cin >> n, n){
int found = 0;
//STRINGの入力
int counter = 0;
for (int i = 0; i < n; i++){
string s;
cin >> s;
len[i] = s.size();
for (int j = 0; j < s.size(); j++){
alphabet[counter++] = s[j];
}
}
//同じ文字の処理
int numcounter = 0, variation = 0;
for (int i = 0; i < counter; i++){
bool flag = true;
for (int j = 0; j < i; j++){
if (alphabet[i] == alphabet[j]){
np[i] = np[j];
flag = false;
}
}
if (flag){
np[i] = num + numcounter++;
variation++;
}
}
int docounter = 0;
int t = fact(10 - variation);
int lketa[8];
int rketa[8];
do{
//重複の場合はスキップ
docounter++;
if (docounter % t != 0)
continue;
//下一桁で判定
numcounter = -1;
int lhito = 0, rhito;
for (int i = 0; i < n - 1; i++){
numcounter = numcounter + len[i];
lhito += *np[numcounter];
}
rhito = *np[numcounter + len[n - 1]];
if (lhito % 10 != rhito)
continue;
//全桁の計算
fill_n(lketa, 8, 0);
fill_n(rketa, 8, 0);
numcounter = 0;
bool flag = false;
for (int i = 0; i < n; i++){
for (int j = 0; j < len[i]; j++){
if (j == 0 && *np[numcounter] == 0 && len[i] > 1){
flag = true;
break;
}
if (i < n - 1)
lketa[len[i] - j - 1] += *np[numcounter++];
else
rketa[len[i] - j - 1] = *np[numcounter++];
}
if (flag)
break;
}
if (flag)
continue;
//繰り上がりと判定
bool judge = true;
for (int i = 0; i < 7; i++){
lketa[i + 1] += lketa[i] / 10;
lketa[i] = lketa[i] % 10;
if (lketa[i] != rketa[i]){
judge = false;
break;
}
}
if (lketa[7] != rketa[7])
judge = false;
if (judge){
found++;
}
} while (next_permutation(num, num + 10));
cout << found << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | #include <bits/stdc++.h>
using namespace std;
map<char,int> m;
vector<int> v = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
set<char> Top;
string S;
int dfs (int n, int sum) {
if (n == m.size()) return sum == 0;
int res = 0;
for (int i = 0; i < v.size(); i++) {
if (v[i] == 0 && Top.find(S[n]) != Top.end()) continue;
int tmp = v[i];
v.erase(v.begin() + i);
res += dfs (n + 1, sum + (m[S[n]] * tmp));
v.insert(v.begin() + i, tmp);
}
return res;
}
int main ()
{
int n;
while (cin >> n, n) {
set<char> s;
m.clear();
Top.clear();
for (int i = 0; i < n; i++) {
string str;
cin >> str;
int k = 1;
if (str.size() > 1) Top.insert(str[0]);
for (int j = str.size()-1, k = 1; j >= 0; j--, k *= 10) {
s.insert(str[j]);
if (m.find(str[j]) == m.end()) {
if (i == n - 1) m[str[j]] = -k;
else m[str[j]] = k;
} else {
if (i == n - 1) m[str[j]] -= k;
else m[str[j]] += k;
}
}
}
S = "";
for (auto i : s) {
S += i;
}
cout << dfs(0, 0) << endl;
}
return 0;
} | CPP |
p00742 Verbal Arithmetic | Let's think about verbal arithmetic.
The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.
905 + 125 = 1030
In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.
ACM + IBM = ICPC
Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.
The rules of this puzzle are the following.
* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.
There are 4 different digit assignments for the verbal equation above as shown in the following table.
Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8
Your job is to write a program which solves this puzzle.
Input
The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.
The number of datasets is no more than 100. Each dataset is formatted as follows.
> N
STRING 1
STRING 2
...
STRING N
The first line of a dataset contains an integer N which is the number of integers appearing in the equation. Each of the following N lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.
Each dataset expresses the following equation.
> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N
The integer N is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.
Output
For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.
The output must not contain any superfluous characters.
Sample Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output for the Sample Input
4
1
8
30
0
0
0
40320
Example
Input
3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0
Output
4
1
8
30
0
0
0
40320 | 7 | 0 | import java.util.*;
public class Main{
int n, cnt, cnum, maxlen;
ArrayList<Character> clist, tlist;
char[][] c;
String[] s;
int[] b;
void check(int[] perm){
int ten = 1;
int suma = 0;
int sumb = 0;
for(int i=0;i<maxlen;i++){
sumb = 0;
for(int j=n-b[i];j<n;j++){
if(j==n-1) sumb -= perm[clist.indexOf(c[j][i])];
else sumb += perm[clist.indexOf(c[j][i])];
}
suma += sumb*ten;
ten *= 10;
if(suma%ten!=0) break;
}
if(suma==0) cnt++;
}
void makeperm(int k, int[] perm, boolean[] flag){
if(k==cnum){
check(perm);
}else{
for(int i=0;i<10;i++){
if(flag[i]==true || (i==0 && tlist.contains(clist.get(k))==true)) continue;
perm[k] = i;
flag[i] = true;
makeperm(k+1, perm, flag);
flag[i] = false;
}
}
}
void solve(){
Scanner sc = new Scanner(System.in);
while(true){
n = sc.nextInt();
if(n==0) break;
s = new String[n];
for(int i=0;i<n-1;i++) s[i] = sc.next();
s[n-1] = "xxxxxxxxxx";
Arrays.sort(s, new Comparator<String>(){
public int compare(String o1, String o2){
return o1.length() - o2.length();
}
});
s[n-1] = sc.next();
if(s[n-1].length()<s[n-2].length() || s[n-1].length()>s[n-2].length()+2){
System.out.println(0);
}else{
clist = new ArrayList<Character>();
tlist = new ArrayList<Character>();
c = new char[n][];
maxlen = s[n-1].length();
b = new int[maxlen];
for(int i=0;i<n;i++){
char[] tmp = s[i].toCharArray();
c[i] = new char[tmp.length];
for(int j=0;j<tmp.length;j++){
if(j==0 && tmp.length>1 && tlist.contains(tmp[j])==false) tlist.add(tmp[j]);
if(clist.contains(tmp[j])==false) clist.add(tmp[j]);
c[i][tmp.length-1-j] = tmp[j];
b[j]++;
}
}
cnum = clist.size();
cnt = 0;
makeperm(0, new int[cnum], new boolean[10]);
System.out.println(cnt);
}
}
}
public static void main(String[] args) {
new Main().solve();
}
} | JAVA |
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