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| description
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| difficulty
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p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <bits/stdc++.h>
#define LL long long
using namespace std;
int N, M, A, B;
int main()
{
scanf("%d%d%d%d", &N, &M, &A, &B);
if (!(N % A) && !(M % B))
puts("No");
else
{
puts("Yes");
for (int i = 0; i < N; ++i)
{
for (int j = 0; j < M; ++j)
{
if (i % A || j % B)
printf("%d ", -1000);
else printf("%d ", 1000 * (A * B - 1) - 1);
}
puts("");
}
}
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
ll INF = 1e9 + 7;
int main()
{
int H, W, h, w;
cin >> H >> W >> h >> w;
if(H % h == 0 && W % w == 0)
{
cout << "No" << endl;
return 0;
}
cout << "Yes" << endl;
for(int i = 1; i <= H; i++)
{
for(int j = 1; j <= W; j++)
{
if(i % h == 0 && j % w == 0) cout << -(h*w-1) * 1000 -1 << " ";
else cout << "1000 ";
}
cout << endl;
}
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int H = sc.nextInt();
int W = sc.nextInt();
long h = sc.nextLong();
long w = sc.nextLong();
long x = (-1000)*(h*w-1)-1;
String[] ans = new String[H];
for(int i=0; i<H; ++i){
ans[i] = "";
}
long sum = 0;
for(int i=1; i<=H; ++i){
StringBuilder sb = new StringBuilder();
for(int j=1; j<=W; j++){
if(i%h==0 &&j%w==0){
sb.append(x);
sb.append(" ");
sum+=x;
}else{
sb.append(1000);
sb.append(" ");
sum+=1000;
}
}
sb.deleteCharAt(sb.length()-1);
ans[i-1] = sb.toString();
}
if(sum<=0) System.out.println("No");
else{
System.out.println("Yes");
for(int i=0; i<H; i++){
System.out.println(ans[i]);
}
}
return;
}
} | JAVA |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
def solve():
H,W,h,w = LI()
if W%w != 0:
s = [0]+[100000]*(w-1)
for i in range(W-w+1):
s.append(s[-w]-1)
a = [s[i]-s[i-1] for i in range(1,W+1)]
print("Yes")
for i in range(H):
print(*a)
elif H%h != 0:
H,W,h,w = W,H,w,h
s = [0]+[100000]*(w-1)
for i in range(W-w+1):
s.append(s[-w]-1)
a = [s[i]-s[i-1] for i in range(1,W+1)]
print("Yes")
ans = [[a[i]]*H for i in range(W)]
for i in range(W):
print(*ans[i])
else:
print("No")
return
#Solve
if __name__ == "__main__":
solve()
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int a[1000][1000],H,W,h,w;
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin>>H>>W>>h>>w;
for (int i=1;i<=H;i++)
for (int j=1;j<=W;j++)
if (i%h==0 && j%w==0) a[i][j]=-696*(h*w-1)-1;
else a[i][j]=696;
int sum=0;
for (int i=1;i<=H;i++)
for (int j=1;j<=W;j++) sum+=a[i][j];
if (sum<=0) {
cout<<"No";
return 0;
}
cout<<"Yes"<<endl;
for (int i=1;i<=H;i++) {
for (int j=1;j<=W;j++) cout<<a[i][j]<<" ";
cout<<endl;
}
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<stdio.h>
int main(){
int H,W,h,w;
scanf("%d%d%d%d",&H,&W,&h,&w);
if(H%h||W%w){
puts("Yes");
for(int i=1;i<=H;++i,puts("")) for(int j=1;j<=W;++j)
if(i%h||j%w) printf("1000 ");
else printf("%d ", -1000*(h*w-1)-1);
}else puts("No");
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
int A[505][505];
int main(){
int H,W,h,w; cin>>H>>W>>h>>w;
if(H%h==0&&W%w==0){
cout<<"No\n";
return 0;
}
ll a=(1e9-1)/(h*w-1);
rep(i,H) rep(j,W){
if(i%h==h-1&&j%w==w-1) A[i][j]=-(a*(h*w-1)+1);
else A[i][j]=a;
}
cout<<"Yes\n";
rep(i,H){
rep(j,W) cout<<A[i][j]<<" ";
cout<<endl;
}
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <cstring>
using namespace std;
const int mod = 1e9 + 7;
int a, b, c, d;
int main() {
cin >> a >> b >> c >> d;
if (a%c == 0 && b%d == 0) { cout << "No" << endl; }
else {
cout << "Yes" << endl;
for (int i = 1; i <= a; ++i) {
for (int j = 1; j <= b; ++j) {
if (i%c == 0 && j%d == 0) { cout << 4000 * (c*d - 1)*-1 - 1 << " "; }
else { cout << "4000 "; }
}
cout << endl;
}
}
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <cstdio>
int H, W, h, w;
int main() {
scanf("%d%d%d%d", &H, &W, &h, &w);
if(H%h + W%w == 0) return !puts("No");
puts("Yes");
for(int i=0; i<H; i++, puts("")) for(int j=0; j<W; j++)
if(H%h) printf("%d ", i%h ? -1000 : 1000*(h-1)-1);
else printf("%d ", j%w ? -1000 : 1000*(w-1)-1);
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<cstdio>
const int MAXN=505;
int H,W,h,w;
int arr[MAXN][MAXN];
int main()
{
scanf("%d%d%d%d",&H,&W,&h,&w);
if(h==1&&w==1)
{
puts("No");
return 0;
}
long long sum=0;
int k=999999999/(h*w-1);
{
sum=0;
for(int i=1;i<=H;i++)
for(int j=1;j<=W;j++)
{
if(i%h==0&&j%w==0)
arr[i][j]=-k*(h*w-1)-1;
else
arr[i][j]=k;
sum+=arr[i][j];
}
if(sum>0)
{
puts("Yes");
for(int i=1;i<=H;i++)
{
for(int j=1;j<=W;j++)
printf("%d ",arr[i][j]);
puts("");
}
return 0;
}
}
puts("No");
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H, W, sh, sw = map(int, input().split())
if H%sh == 0 and W%sw == 0:
print("No")
else:
print("Yes")
state = [[1]*W for _ in range(H)]
INF = 10**9
for h in range(H):
for w in range(W):
if w%sw == sw-1 and h%sh == sh-1:
state[h][w] = -INF
elif w%sw == 0 and h%sh == 0:
state[h][w] = INF - (sh*sw-1)
for h in range(H):
print(" ".join([str(a) for a in state[h]]))
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #!/usr/bin/env python3
H, W, h, w = map(int, input().split())
if W % w == 0 and H % h == 0:
print('No')
else:
swapped = False
if W % w == 0:
H, W, h, w, swapped = W, H, w, h, True
a = [ [ 0 for _ in range(W) ] for _ in range(H) ]
for y in range(H):
for x in range(0, W, w):
a[y][x] = W
for x in range(w - 1, W, w):
a[y][x] = - W - 1
if swapped:
H, W, h, w = W, H, w, h
b = a
a = [ [ 0 for _ in range(W) ] for _ in range(H) ]
for y in range(H):
for x in range(W):
a[y][x] = b[x][y]
print('Yes')
for y in range(H):
print(*a[y])
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H,W,h,w=[int(i) for i in input().split()]
if H%h==0 and W%w==0:
print("No")
elif W%w!=0:
print("Yes")
for y in range(H):
for x in range(W):
if x!=0:
print(" ",end="")
if (x%w)!=(w-1):
print(100000,end="")
else:
print(-100000*(w-1)-1,end="")
print()
else:
print("Yes")
for y in range(H):
for x in range(W):
if x!=0:
print(" ",end="")
if (y%h)!=(h-1):
print(100000,end="")
else:
print(-100000*(h-1)-1,end="")
print() | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H,W,h,w = map(int,input().split())
if not (H%h or W%w):
print("No")
exit()
if H%h:
flg = "tate"
n = H
r = h
else:
flg = "yoko"
n = W
r = w
ans = []
if r%2 == 0:
s = r//2
for i in range(n):
if (i//s)%2 == 0:
ans.append(2000)
else:
ans.append(-2001)
else:
for i in range(n):
if (i%r) < r//2+1:
ans.append(2000)
else:
ans.append(-(int(2000/(r//2)*(r//2+1)))-1)
print("Yes")
if flg == "tate":
for i in range(H):
print(*([ans[i]]*W))
else:
for i in range(H):
print(*ans) | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int main()
{
int H, W, h, w;
cin >> H >> W >> h >> w;
int nrlin = H / h;
int nrcol = W / w;
if (H % h == 0 && W % w == 0) {
cout << "No\n";
return 0;
}
else
cout << "Yes\n";
for (int i(1); i <= H; i++)
for (int j(1); j <= W; j++)
cout << (i % h == 0 && j % w == 0 ? -(h * w - 1) * 1000 - 1 : 1000) << " \n"[j == W];
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import sys
input = sys.stdin.readline
def main():
H,W,h,w = map(int,input().split())
if H*W-(H//h)*(W//w)*w*h == 0:
print("No")
else:
if W%w != 0:
ans = [0 for _ in range(W)]
co = [0 for _ in range(W+1)]
for i in range(1,w):
co[i] = (W-i)//w+1
for i in range(w,W+1):
co[i] = co[i-w]-1
for i in range(W):
ans[i] = co[i+1]-co[i]
print("Yes")
for i in range(H):
print(*ans)
else:
ans = [0 for _ in range(H)]
co = [0 for _ in range(H+1)]
for i in range(1,h):
co[i] = (H-i)//h+1
for i in range(h,H+1):
co[i] = co[i-h]-1
for i in range(H):
ans[i] = co[i+1]-co[i]
print("Yes")
for i in range(H):
print(*list(ans[i] for _ in range(W)))
if __name__ == "__main__":
main() | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H,W,h,w=map(int,input().split())
if H%h==0 and W%w==0:
print("No")
else:
print("Yes")
for i in range(H):
print(' '.join("-1000" if i%h or j%w else str(1000*(h*w-1)-1) for j in range(W))) | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<stdio.h>
int main(){int H,W,h,w,i=1,j;scanf("%d%d%d%d",&H,&W,&h,&w);if(H%h||W%w)for(puts("Yes");i<=H;++i,puts(""))for(j=0;j<W;++j,printf("%d ",(i%h||j%w)?999:-999*(h*w-1)-1));else puts("No");return 0;} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import java.io.*;
import java.util.Arrays;
import java.util.StringJoiner;
import java.util.StringTokenizer;
public class Main {
static int H, W, h, w;
public static void main(String[] args) {
FastScanner sc = new FastScanner(System.in);
H = sc.nextInt();
W = sc.nextInt();
h = sc.nextInt();
w = sc.nextInt();
int[][] ans = solve();
if( ans != null ) {
PrintWriter pw = new PrintWriter(System.out);
pw.println("Yes");
for (int i = 0; i < H; i++) {
StringJoiner sj = new StringJoiner(" ");
for (int j = 0; j < W; j++) {
sj.add( String.valueOf(ans[i][j]) );
}
pw.println(sj.toString());
}
pw.flush();
} else {
System.out.println("No");
}
}
static int[][] solve() {
// 負できっちり被覆されるのでどうにもならない
if( H % h == 0 && W % w == 0 ) return null;
// 倍数ではない行列毎に同じことをしても問題ない
// aaabaaaba のように構成するとして aaab = -1となるように置く
// aが十分に大きいなら全体として正になる
int[][] ans = new int[H][W];
if( H % h == 0 ) {
int a = 1000;
int b = -(1000*(w-1) + 1);
for (int i = 0; i < H; i++) {
for (int j = 0; j < W; j++) {
if( j % w == w-1 ) {
ans[i][j] = b;
} else {
ans[i][j] = a;
}
}
}
} else {
int a = 1000;
int b = -(1000*(h-1) + 1);
for (int i = 0; i < W; i++) {
for (int j = 0; j < H; j++) {
if( j % h == h -1 ) {
ans[j][i] = b;
} else {
ans[j][i] = a;
}
}
}
}
// int sum = 0;
// for (int i = 0; i < H; i++) {
// for (int j = 0; j < W; j++) {
// sum += ans[i][j];
// }
// }
// debug(sum);
return ans;
}
@SuppressWarnings("unused")
static class FastScanner {
private BufferedReader reader;
private StringTokenizer tokenizer;
FastScanner(InputStream in) {
reader = new BufferedReader(new InputStreamReader(in));
tokenizer = null;
}
String next() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
String nextLine() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
return reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken("\n");
}
long nextLong() {
return Long.parseLong(next());
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
int[] nextIntArray(int n, int delta) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt() + delta;
return a;
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
}
static void writeLines(int[] as) {
PrintWriter pw = new PrintWriter(System.out);
for (int a : as) pw.println(a);
pw.flush();
}
static void writeLines(long[] as) {
PrintWriter pw = new PrintWriter(System.out);
for (long a : as) pw.println(a);
pw.flush();
}
static void writeSingleLine(int[] as) {
PrintWriter pw = new PrintWriter(System.out);
for (int i = 0; i < as.length; i++) {
if (i != 0) pw.print(" ");
pw.print(i);
}
pw.println();
pw.flush();
}
static int max(int... as) {
int max = Integer.MIN_VALUE;
for (int a : as) max = Math.max(a, max);
return max;
}
static int min(int... as) {
int min = Integer.MAX_VALUE;
for (int a : as) min = Math.min(a, min);
return min;
}
static void debug(Object... args) {
StringJoiner j = new StringJoiner(" ");
for (Object arg : args) {
if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg));
else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg));
else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg));
else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg));
else j.add(arg == null ? "null" : arg.toString());
}
System.err.println(j.toString());
}
static void printSingleLine(int[] array) {
for (int i = 0; i < array.length; i++) {
if (i != 0) System.out.print(" ");
System.out.print(array[i]);
}
System.out.println();
}
static int lowerBound(int[] array, int value) {
int low = 0, high = array.length, mid;
while (low < high) {
mid = ((high - low) >>> 1) + low;
if (array[mid] < value) low = mid + 1;
else high = mid;
}
return low;
}
static int upperBound(int[] array, int value) {
int low = 0, high = array.length, mid;
while (low < high) {
mid = ((high - low) >>> 1) + low;
if (array[mid] <= value) low = mid + 1;
else high = mid;
}
return low;
}
}
| JAVA |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <iostream>
using namespace std;
int main() {
int H, W, h, w;
cin >> H >> W >> h >> w;
if(H%h<1 && W%w<1) cout << "No" << endl;
else{
cout << "Yes" << endl;
int T=((H/h)*(W/w))/(W*(H%h)+H*(W%w)-(H%h)*(W%w))+1;
for(int i=1; i<=H; ++i){
for(int j=1; j<=W; ++j){
if(i%h<1 && j%w<1) cout << -T*(h*w-1)-1;
else cout << T;
if(j<W) cout << ' ';
else cout << endl;
}
}
}
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
#define ll long long
using namespace std;
ll H, W, h, w;
int main() {
cin >> H >> W >> h >> w;
if (H % h == 0 && W % w == 0) return printf("No"), 0;
puts("Yes");
if (H % h != 0) {
for (int i = 0; i < H; i ++) {
for (int j = 0; j < W; j ++)
if (i % h == 0) printf("%lld ", 1000 * h - 1001);
else printf("-1000 ");
puts("");
}
} else {
for (int i = 0; i < H; i ++) {
for (int j = 0; j < W; j ++)
if (j % w == 0) printf("%lld ", 1000 * w - 1001);
else printf("-1000 ");
puts("");
}
}
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H, W, h, w = map(int, input().split())
if H % h == 0 and W % w == 0:
print("No")
exit(0)
blocks = (H//h) * (W//w)
left = H * W - blocks * h * w
offset = blocks // left + 1
v = [offset, -offset * (h * w - 1) - 1]
print("Yes")
for i in range(H):
a = []
for j in range(W):
a.append(v[1 if i % h == h-1 and j % w == w-1 else 0])
print(" ".join(list(map(str, a))))
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int H, W, h, w, a[505][505];
long long ret;
int main () {
scanf("%d%d%d%d", &H, &W, &h, &w);
for (int i = 1; i <= H; i += h) for (int j = 1; j <= W; j += w) {
a[i][j] = 1e9 - 1;
ret += 1e9 - 1;
}
for (int i = h; i <= H; i += h) for (int j = w; j <= W; j += w) {
a[i][j] = -1e9;
ret -= 1e9;
}
if (ret <= 0) return puts("No"), 0;
puts("Yes");
for (int i = 1; i <= H; ++i) for (int j = 1; j <= W; ++j) printf("%d%c", a[i][j], j == W ? '\n' : ' ');
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
const int x=1000;
int W,H,w,h,sum;
int a[510][510];
int main(){
scanf("%d%d%d%d",&W,&H,&w,&h);
for (int i=1;i<=W;++i)
for (int j=1;j<=H;++j)
a[i][j]=x;
sum=W*H*x;
for (int i=w;i<=W;i+=w)
for (int j=h;j<=H;j+=h){
a[i][j]=-w*h*x+x-1;
sum=sum-w*h*x-1;
}
if (sum<=0) puts("No");
else{
puts("Yes");
for (int i=1;i<=W;++i){
for (int j=1;j<=H;++j) printf("%d ",a[i][j]);
puts("");
}
}
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int main(){
int n,m,h,w;
scanf("%d%d%d%d",&n,&m,&h,&w);
int k=1000000000/h/w;
long long ans=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if(i%h==0&&j%w==0) ans-=(h*w-1)*k+1;
else ans+=k;
if(ans<=0) puts("No");
else{
puts("Yes");
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++)
if(i%h==0&&j%w==0) printf("%d ",-(h*w-1)*k-1);
else printf("%d ",k);
putchar('\n');
}
}
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<cstdio>
int main()
{
int n,m,h,w;
scanf("%d%d%d%d",&n,&m,&h,&w);
if( n%h==0&&m%w==0 )
return puts("No"),0;
puts("Yes");
long long c=(n/h)*(m/w)/((n%h)*m+(m%w)*n-(n%h)*(m%w))+1;
long long s=(h*w-1)*c+1,t=0;int i,j;
for(i=1;i<=n;i++,putchar(10))
for(j=1;j<=m;j++)
printf("%lld ",i%h==0&&j%w==0?-s:c),t+=i%h==0&&j%w==0?-s:c;
if(t<=0)return 1;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H,W,h,w=map(int,input().split())
if H%h==0 and W%w==0:
print("No")
else:
c=10**9
ans=[[(c-1)//(h*w-1) for i in range(W)] for j in range(H)]
for i in range(H):
for j in range(W):
if (i+1)%h==0 and (j+1)%w==0:
ans[i][j]=-c
if sum(ans[i][j] for i in range(H) for j in range(W))>0:
print("Yes")
for i in range(H):
print(*ans[i])
else:
print("No")
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H, W, h, w = map(int, input().split())
a = [[0] * W for i in range(H)]
INF = 10**9
for i in range(H):
for j in range(W):
if i % h == h - 1 and j % w == w - 1:
a[i][j] = -INF
elif i % h == 0 and j % w == 0:
a[i][j] = INF - 1
if sum([sum(line) for line in a]) > 0:
print("Yes")
for i in range(H):
print(*a[i])
else:
print("No") | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | H, W, a, b = map(int, input().split())
if H%a or W%b:
print('Yes')
k = (H//a) * (W//b)
x = k+2
ansss = [[0]*(W) for _ in range(H)]
for i in range(H//a):
for j in range(W//b):
ansss[(i+1)*a-1][(j+1)*b-1] = -x
for i in range(-(-H//a)):
for j in range(-(-W//b)):
ansss[i*a][j*b] = x-1
print('\n'.join([' '.join(map(str, anss)) for anss in ansss]))
else:
print('No')
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include "bits/stdc++.h"
using namespace std;
int main() {
int H, W, h, w;
cin >> H >> W >> h >> w;
if (H % h == 0 && W % w == 0) {
cout << "No" << endl;
return 0;
}
cout << "Yes" << endl;
for (int i = 1; i <= H; i++) {
for (int j = 1; j <= W; j++) {
if (i % h == 0 && j % w == 0) cout << -1 * h * w * 250 + 249;
else cout << 250;
if (j == W) cout << endl;
else cout << " ";
}
}
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<stdio.h>
int main(){
int H, W, h, w;
scanf("%d%d%d%d", &H,&W,&h,&w);
if(H%h||W%w){
puts("Yes");
for(int i=0; i<H; ++i,puts(""))
for(int j=0; j<W; ++j)
if((i+1)%h || (j+1)%w) printf("1000 ");
else printf("%d ", -1000*(h*w-1)-1);
}else puts("No");
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int H,W,h,w;
int main()
{
scanf("%d%d",&H,&W);
scanf("%d%d",&h,&w);
if(!(W % w) && !(H % h))
{
printf("No\n");
return 0;
}
printf("Yes\n");
int base,exp;
base = (H / h) * (W / w) + 1;
exp = -1 * (base * (h * w - 1) + 1);
for(int i = 1; i <= H; i++)
{
for(int j = 1; j <= W; j++)
{
if(i % h == 0 && j % w == 0)printf("%d",exp);
else printf("%d",base);
if(j != W)printf(" ");
}
printf("\n");
}
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int main(){
int H,W,h,w;
cin>>H>>W>>h>>w;
if(H%h==0 && W%w==0)
cout<<"No"<<endl;
else{
cout<<"Yes"<<endl;
vector<vector<int>>A(H,vector<int>(W));
for(int i=0;i<H;i++)
for(int j=0;j<W;j++)
if(i%h==h-1 && j%w==w-1)
A[i][j]=-4000*(h*w-1)-1;
else
A[i][j]=4000;
for(int i=0;i<H;i++){
for(int j=0;j<W-1;j++)
cout<<A[i][j]<<" ";
cout<<A[i][W-1]<<endl;
}
}
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
int main() {
int A,B,a,b;
cin>>A>>B>>a>>b;
if(A%a==0&&B%b==0) return puts("No"),0;
int k=999999999/(a*b-1);
puts("Yes");
for(int i=1;i<=A;i++) {
for(int j=1;j<=B;j++) {
if(i%a==0&&j%b==0) printf("%d ",-(a*b-1)*k-1);
else printf("%d ",k);
}
puts("");
}
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | def get_1d(H, h):
ret = [0] * (H + 1)
for s in range(h):
for x, i in enumerate(range(s, H, h)):
ret[i] = -x
for x, i in enumerate(range(H, 0, -h)):
ret[i] = x + 1
return [x1 - x0 for x0, x1 in zip(ret, ret[1:])]
def solve(H, W, h, w):
if H % h == 0 and W % w == 0:
return False
ans = []
if H % h > 0:
col = get_1d(H, h)
ans.extend([x] * W for x in col)
else:
row = get_1d(W, w)
ans.extend([row] * H)
return ans
H, W, h, w = map(int, input().split())
ans = solve(H, W, h, w)
if ans == False:
print('No')
else:
print('Yes')
for row in ans:
print(*row)
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import sys
input = sys.stdin.readline
sys.setrecursionlimit(pow(10, 6))
def main():
H, W, h, w = map(int, input().split())
if H % h == 0 and W % w == 0:
print("No")
elif H % h == 0:
print("Yes")
ans = [[0 for _ in range(W)] for _ in range(H)]
for i in range(H):
for j in range(0, W, w):
ans[i][j] = 99999
if j + w - 1 < W:
ans[i][j + w - 1] = -100000
for a in ans:
print(*a)
else:
print("Yes")
ans = [[0 for _ in range(W)] for _ in range(H)]
for i in range(W):
for j in range(0, H, h):
ans[j][i] = 99999
if j + h - 1 < H:
ans[j + h - 1][i] = -100000
for a in ans:
print(*a)
if __name__ == '__main__':
main()
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int H = in.nextInt();
int W = in.nextInt();
int h = in.nextInt();
int w = in.nextInt();
if (H % h == 0 && W % w == 0) {
out.println("No");
} else {
out.println("Yes");
for (int r = 0; r < H; ++r) {
for (int c = 0; c < W; ++c) {
if (r % h == 0 && c % w == 0) {
out.print((int) 1e9 - 1);
} else if (r % h == h - 1 && c % w == w - 1) {
out.print((int) -1e9);
} else {
out.print(0);
}
out.print(" ");
}
out.println();
}
}
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| JAVA |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
H,W,h,w = LI()
f = False
if H % h == 0 and W % w == 0:
return 'No'
if W % w == 0:
H,W,h,w = W,H,w,h
f = True
p = 10**6*(w-1) - 1
m = -(10**6)
a = []
for i in range(W):
if i % w == 0:
a.append(p)
else:
a.append(m)
aa = [a] * H
r = aa
if f:
r = [[aa[i][j] for i in range(H)] for j in range(W)]
return 'Yes\n' + '\n'.join(' '.join(map(str,t)) for t in r)
print(main())
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int H, W, HS, WS;
#define BIG 4000
int main() {
cin >> H >> W >> HS >> WS;
if(H % HS == 0 && W % WS == 0) cout << "No" << endl;
else {
cout << "Yes" << endl;
for(int i = 0; i < H; ++i) {
for(int j = 0; j < W; ++j) {
cout << (j == 0 ? "" : " ")
<< (i % HS == HS - 1 && j % WS == WS - 1 ? BIG - 1 - BIG * HS * WS : BIG);
}
cout << endl;
}
}
return 0;
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
const int N=505;
int H,W,h,w,i,j,a[N][N],s;
int main(){
scanf("%d%d%d%d",&H,&W,&h,&w);
for(i=0;i<H;++i)for(j=0;j<W;++j)s+=a[i][j]=i%h==h-1 && j%w==w-1?-h*w-(1<<20):
(i%h==0 && j%w==0?h*w+(1<<20)-1:0);
if(s<0){
puts("No");
return 0;
}
puts("Yes");
for(i=0;i<H;++i)for(j=0;j<W;++j)printf("%d%c",a[i][j],"\n "[j<W-1]);
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | h, w, h0, w0 = map(int, input().split())
if h0 == w0 == 1:
print("No")
exit()
x, y = h // h0, w // w0
a1cnt = h0 * w0 - 1
a1sum, a2sum = h * w - x * y, x * y
a1 = ((pow(10, 9) - 1) // a1cnt)
a2 = -(a1 * a1cnt + 1)
asum = a1 * a1sum + a2 * a2sum
print("Yes" if asum > 0 else "No")
if asum > 0:
for i in range(1, h + 1):
a = [0] * w
for j in range(1, w + 1):
a[j - 1] = a2 if i % h0 == j % w0 == 0 else a1
print(*a) | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int n,m,h,w;
int main()
{
scanf("%d%d%d%d",&n,&m,&h,&w);
if(n%h||m%w)
{
puts("Yes");
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(i%h||j%w)printf("1000 ");else printf("%d ",-1000*(h*w-1)-1);
}
puts("");
}
}
else puts("No");
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | # coding: utf-8
import array, bisect, collections, copy, heapq, itertools, math, random, re, string, sys, time
sys.setrecursionlimit(10 ** 7)
INF = 10 ** 20
MOD = 10 ** 9 + 7
def II(): return int(input())
def ILI(): return list(map(int, input().split()))
def IAI(LINE): return [ILI() for __ in range(LINE)]
def IDI(): return {key: value for key, value in ILI()}
def read():
H, W, h, w = ILI()
return H, W, h, w
def solve(H, W, h, w):
if H % h == 0 and W % w == 0:
return False, None
else:
matrix = [[10 ** 3] * W for __ in range(H)]
val = - (h * w * 1000) + 999
for _h in range(h - 1, H, h):
for _w in range(w - 1, W, w):
matrix[_h][_w] = val
return True, matrix
def main():
params = read()
ans_bool, matrix = solve(*params)
if ans_bool:
print("Yes")
for row in matrix:
print(" ".join(map(str, row)))
else:
print("No")
if __name__ == "__main__":
main()
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | import sys
def MI(): return map(int,sys.stdin.readline().rstrip().split())
H,W,h,w = MI()
q1,r1 = H//h,H % h
q2,r2 = W//w,W % w
if r1 == 0 and r2 == 0:
print('No')
else:
print('Yes')
if r2 != 0:
A = []
for _ in range(q2):
A += [2*10**6]*(w-1)+[-(2*10**6)*(w-1)-1]
A += [2*10**6]*r2
for _ in range(H):
print(*A)
else:
for i in range(1,H+1):
if i % h != 0:
print(*([2*10**6]*W))
else:
print(*([-(2*10**6)*(h-1)-1]*W))
| PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(int i=l;i<=r;++i)
const int K=500+5;
int main()
{
//freopen("1.in","r",stdin);
int H,W,h,w;
cin>>H>>W>>h>>w;
if(H%h==0&&W%w==0)
{
puts("No");
exit(0);
}
puts("Yes");
int v=-(((h*w)-1)*K+1);
rep(i,1,H)
{
rep(j,1,W)
printf("%d ",(i%h==0&&j%w==0)?v:K);
puts("");
}
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<stdio.h>
int main(){
int H, W, h, w;
scanf("%d%d%d%d", &H,&W,&h,&w);
if(H%h==0 && W%w==0){
puts("No");
return 0;
}
puts("Yes");
for(int i=0; i<H; ++i,puts(""))
for(int j=0; j<W; ++j)
if(i%h==h-1 && j%w==w-1) printf("-999999 ");
else if(i%h==0 && j%w==0) printf("999998 ");
else printf("0 ");
return 0;
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int H, W, h, w;
int main() {
scanf("%d %d %d %d", &H, &W, &h, &w);
if(H % h == 0 && W % w == 0) {
printf("No");
return 0;
}
printf("Yes\n");
int q1 = H / h;
int q2 = W / w;
int x = (q1 + 1) * (q2 + 1);
int y = h*w*x - x - 1;
for(int i = 0; i < H; i++) {
for(int j = 0; j < W; j++) {
if(i % h == 0 && j % w == 0) printf("%d ", y);
else printf("%d ", -x);
}
printf("\n");
}
}
| CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include <iostream>
using namespace std;
int H,W,h,w;
int main() {
cin >> H >> W >> h >> w;
if (H%h != 0 || W%w != 0) {
cout << "Yes\n";
int k=(H/h)*(W/w)+1;
for (int i=0; i<H; i++) {
for (int j=0; j<W; j++) {
if (i%h == h-1 && j%w == w-1)
cout << -k-1 << " ";
else if (i%h == 0 && j%w == 0)
cout << k << " ";
else
cout << "0 ";
}
cout << "\n";
}
} else {
cout << "No\n";
}
} | CPP |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #!/usr/bin/env python3
import sys
try: from typing import Any, Union, List, Tuple, Dict
except ImportError: pass
def debug(*args): print(*args, file=sys.stderr)
def exit(): sys.exit(0)
H, W, h, w = map(int, input().split())
yes, no = "Yes", "No"
if (H%h == 0) and (W%w == 0):
print(no)
exit()
print(yes)
unit = 1000
if (W%w != 0):
for i in range(H):
# l_ = [str(unit)] * (w-1)
# l_ += [str(-unit*(w-1)-1)]
# l = l_ * (W//w)
# l += [str(unit)] * (W%w)
l = []
for j in range(W):
if j % w == w-1:
l.append(str(-unit*(w-1)-1))
else:
l.append(str(unit))
print(" ".join(l))
exit()
if (H%h != 0):
for i in range(H):
if i%h == h-1:
l = [str(-unit*(h-1)-1)] * W
else:
l = [str(unit)] * W
print(" ".join(l))
exit()
raise Exception("hogeeee") | PYTHON3 |
p03689 AtCoder Grand Contest 016 - +/- Rectangle | You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W). Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:
* The matrix has H rows and W columns.
* Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).
* The sum of all the elements of the matrix is positive.
* The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.
Constraints
* 1 ≤ h ≤ H ≤ 500
* 1 ≤ w ≤ W ≤ 500
Input
Input is given from Standard Input in the following format:
H W h w
Output
If there does not exist a matrix that satisfies all of the conditions, print `No`.
Otherwise, print `Yes` in the first line, and print a matrix in the subsequent lines in the following format:
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}
Here, a_{ij} represents the (i,\ j) element of the matrix.
Examples
Input
3 3 2 2
Output
Yes
1 1 1
1 -4 1
1 1 1
Input
2 4 1 2
Output
No
Input
3 4 2 3
Output
Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define FOR(i,x,y) for(int i=(int)x; i<(int)y; ++i)
int main(){
int H, W, h, w;
cin >> H >> W >> h >> w;
if(H%h==0 && W%w==0){
printf("No\n");
}else{
printf("Yes\n");
FOR(i,0,H){
FOR(j,0,W){
if(i%h==h-1 && j%w==w-1) printf("-999999 ");
else if(i%h==0 && j%w==0) printf("999998 ");
else printf("0 ");
}
printf("\n");
}
}
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define mo 1000000007ll
#define N 100010
#define int long long
int n,d[N],a[N],vi[N],c[N],le[N],s[N],f[N],ans;
void dfs(int x){
int v=0,y=0,z=0,p=0;
while(c[x]){
v++;
c[x]=0;
if(le[x]){
if(!y){
z=y=v;
p=le[x];
}
else{
ans=ans*((le[x]<v-z)+(le[x]<=v-z))%mo;
z=v;
}
}
x=a[x];
}
if(!y)s[v]++;
else ans=ans*((p<v-z+y)+(p<=v-z+y))%mo;
}
void dp(){
for(int i=1;i<=n;i++){
if(d[i])continue;
int x=i,l=0;
while(!c[x]){
x=a[x];
l++;
}
le[x]=l;
}
ans=1;
for(int i=1;i<=n;i++)
if(c[i])dfs(i);
for(int i=1;i<=n;i++)
if(s[i]){
f[0]=1;
for(int j=1;j<=s[i];j++){
if(i>1&&(i&1))
f[j]=f[j-1]*2ll%mo;
else f[j]=f[j-1];
if(j>1)f[j]=(f[j]+f[j-2]*(j-1)%mo*i%mo)%mo;
}
ans=ans*f[s[i]]%mo;
}
}
signed main(){
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
d[a[i]]++;
}
for(int i=1;i<=n;i++){
if(vi[i])continue;
int x=i;
while(!vi[x]){
vi[x]=i;
x=a[x];
}
if(vi[x]!=i)continue;
while(!c[x]){
c[x]=1;
x=a[x];
}
}
for(int i=1;i<=n;i++)
if((c[i]&&d[i]>2)||(!c[i]&&d[i]>1)){
puts("0");
return 0;
}
dp();
cout<<ans;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
typedef long long LL;
const int Mod = 1000000007;
const int MN = 100005;
int N, p[MN], c[MN], d[MN], k[MN], g[MN], Ans;
int q[MN], l, r;
int main() {
scanf("%d", &N);
for (int i = 1; i <= N; ++i) scanf("%d", &p[i]), ++d[p[i]];
for (int i = 1; i <= N; ++i) if (d[i] > 2) return puts("0"), 0;
for (int i = 1; i <= N; ++i) if (!d[i]) q[++r] = i;
while (l < r) {
int u = q[++l], x = p[u];
k[x] = k[u] + 1;
if (++c[x] == d[x]) q[++r] = x;
}
for (int i = 1; i <= N; ++i) if (c[i] > 1) return puts("0"), 0;
Ans = 1;
for (int i = 1; i <= N; ++i) if (c[i] < d[i]) {
q[r = 1] = i;
for (int x = p[i]; x != i; x = p[x]) q[++r] = x;
int lst = 0;
for (int j = 1; j <= r; ++j) if (c[q[j]]) lst = j;
for (int j = 1; j <= r; ++j) c[q[j]] = d[q[j]];
if (!lst) { ++g[r]; continue; }
lst -= r;
for (int j = 1; j <= r; ++j) if (k[q[j]]) {
if (k[q[j]] > j - lst) return puts("0"), 0;
if (k[q[j]] < j - lst) Ans = 2 * Ans % Mod;
lst = j;
}
}
for (int i = 1; i <= N; ++i) if (g[i]) {
int x = 1, y = 1 + (i >= 3 && i & 1), z;
for (int j = 2; j <= g[i]; ++j)
z = ((1 + (i >= 3 && i & 1)) * y + (LL)x * (j - 1) * i) % Mod, x = y, y = z;
Ans = (LL)Ans * y % Mod;
}
printf("%d\n", Ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<vector>
#define N 100100
#define Q 1000000007
using namespace std;
bool vis[N];
int a[N],stk[N],cnt,num[N],cc,lf[N];
vector<int> cyc[N],g[N];
int dp[N];
int main(){
int n,ans=1;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),g[a[i]].push_back(i);
for(int i=1;i<=n;i++){
if(!vis[i]){
int j;
cnt=0;
for(j=i;!vis[j];j=a[j]){
vis[j]=true;
stk[cnt++]=j;
}
do{
cyc[cc].push_back(stk[--cnt]);
}while(cnt&&stk[cnt]!=j);
if(stk[cnt]==j) cc++;
else cyc[cc].clear();
}
}
for(int i=0;i<cc;i++){
bool fx=false;
for(int j=0;j<cyc[i].size();j++){
int u=cyc[i][j];
if(g[u].size()>2) ans=0;
else if(g[u].size()==2){
fx=true;
lf[j]=0;
int v=g[u][0]^g[u][1]^(cyc[i][(j+1)%cyc[i].size()]);
while(1){
lf[j]++;
if(g[v].size()==1) v=g[v][0];
else{
if(g[v].size()>1) ans=0;
break;
}
}
}
else lf[j]=0;
}
if(!fx) num[cyc[i].size()]++;
else{
for(int j=0;j<cyc[i].size();j++){
if(lf[j]){
for(int k=(j+1)%cyc[i].size();g[cyc[i][k]].size()==1;k=(k+1)%cyc[i].size()){
lf[j]--;
}
if(lf[j]<=0) ans=(ans+ans)%Q;
else if(lf[j]>1) ans=0;
}
}
}
}
for(int i=1;i<=n;i++){
dp[0]=1;
dp[1]=1+(i>1)*(i&1);
for(int j=2;j<=num[i];j++){
dp[j]=(dp[1]*dp[j-1]+1LL*(j-1)*i%Q*dp[j-2])%Q;
}
ans=1LL*ans*dp[num[i]]%Q;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=100005,M=1e9+7;
int n,a[N],pre[N],deg[N],cnt[N],f[N],ans=1,vis[N];
signed main(){
scanf("%lld",&n);
for (int i=1;i<=n;i++){
scanf("%lld",&a[i]);
deg[a[i]]++;
}
for (int i=1;i<=n;i++){
if (deg[i]>2){
printf("0\n");
return 0;
}
if (deg[i]<2||vis[i])continue;
int p=i;
while (p!=i||!vis[p]){
if (vis[p]){
printf("0\n");
return 0;
}
vis[p]=1;
pre[a[p]]=p;
p=a[p];
}
}
for (int i=1;i<=n;i++)
if (!deg[i]){
int p=i,l1=0,l2=0;
while (!vis[p])vis[p]=1,p=a[p],l1++;
do {
l2++;
p=pre[p];
}while(deg[p]!=2);
if (l1<l2)ans=ans*2%M;
else if (l1>l2){
printf("0\n");
return 0;
}
}
for (int i=1;i<=n;i++)
if (!vis[i]){
int p=i,l=0;
do {
l++;
p=a[p];
vis[p]=1;
}
while(p!=i);
cnt[l]++;
}
for (int i=1;i<=n;i++){
int mul=1;
if (i!=1&&(i&1))mul++;
f[0]=1;f[1]=mul;
for (int j=2;j<=cnt[i];j++)
f[j]=(f[j-2]*(j-1)%M*i+f[j-1]*mul)%M;
ans=ans*f[cnt[i]]%M;
}
printf("%lld\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
const int MAXN = 100010;
const int mod = 1000000007;
typedef long long LL;
void bye() { std::cout << 0 << std::endl; exit(0); }
void reduce(int & x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (LL) a * b % mod; }
int pow(int a, int b, int res = 1) {
for (; b; b >>= 1, a = mul(a, a)) if (b & 1) res = mul(res, a);
return res;
}
int A[MAXN], n, ind[MAXN], vis[MAXN];
int fac[MAXN], inv[MAXN], cnt[MAXN];
int C(int a, int b) { return (LL) fac[a] * inv[b] % mod * inv[a - b] % mod; }
std::vector<int> in[MAXN];
int ans = 1;
int main() {
std::ios_base::sync_with_stdio(false), std::cin.tie(0);
fac[0] = inv[0] = fac[1] = inv[1] = 1;
for (int i = 2; i != MAXN; ++i) {
fac[i] = mul(fac[i - 1], i);
inv[i] = mul(inv[mod % i], mod - mod / i);
}
for (int i = 2; i != MAXN; ++i)
inv[i] = mul(inv[i - 1], inv[i]);
std::cin >> n;
for (int i = 1; i <= n; ++i)
std::cin >> A[i], ++ind[A[i]], in[A[i]].push_back(i);
for (int i = 1; i <= n; ++i) if (ind[i] > 2) bye();
for (int i = 1; i <= n; ++i) if (!vis[i] && !ind[i]) {
static int li[MAXN], bak, cir[MAXN]; bak = 0;
auto pre = [] (int x) { return x == 1 ? bak : x - 1; };
int now = i;
while (!vis[now]) {
vis[now] = true;
li[++bak] = now;
now = A[now];
}
int at = std::find(li + 1, li + 1 + bak, now) - li;
bak -= at - 1;
for (int i = 1; i <= bak; ++i) li[i] = li[i + at - 1];
for (int i = 1; i <= bak; ++i) {
int u = li[i]; cir[i] = 0;
for (auto t : in[u]) if (t != li[pre(i)]) {
int now = t;
while (true) {
++cir[i];
if (in[now].size() > 1) bye();
if (in[now].empty()) break;
now = in[now][0];
}
}
}
for (int i = 1; i <= bak; ++i) if (cir[i]) {
int dis = 1, now = pre(i);
while (!cir[now]) now = pre(now), ++dis;
if (dis < cir[i]) bye();
ans = mul(ans, 1 + (dis > cir[i]));
}
}
for (int i = 1; i <= n; ++i) if (!vis[i]) {
int now = A[i], l = 1; vis[i] = true;
while (now != i) ++l, vis[now] = true, now = A[now];
++cnt[l];
}
for (int i = 1; i <= n; ++i) {
int t = 0;
for (int j = 0; j <= cnt[i]; j += 2) {
int t2 = (LL) C(cnt[i], j) * C(j, j / 2) % mod * fac[j / 2] % mod;
t2 = pow(mul(i, mod + 1 >> 1), j / 2, t2);
if (i > 1 && i % 2 == 1)
t2 = pow(2, cnt[i] - j, t2);
reduce(t += t2 - mod);
}
ans = mul(ans, t);
}
std::cout << ans << std::endl;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
int q=0;char ch=' ';
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
return q;
}
const int mod=1e9+7,N=100005;
int n,ans;
int a[N],du[N],cir[N],vis[N],footL[N],sum[N],f[N];
int qm(int x) {return x>=mod?x-mod:x;}
void workcir(int x) {
int now=0,fr=0,ed=0,frL=0;
//fr:第一个有脚的位置，ed:上一个找到的有脚的位置
//frL:第一个脚的长度，now:当前节点是从x开始走环走到的第几个点
while(cir[x]) {
++now,cir[x]=0;
if(footL[x]) {
if(!fr) ed=fr=now,frL=footL[x];
else {//塞脚
int kl=(footL[x]<now-ed)+(footL[x]<=now-ed);
ans=1LL*ans*kl%mod,ed=now;
}
}
x=a[x];
}
if(!fr) ++sum[now];//是简单环
else {//考虑第一个脚
int kl=(frL<now-ed+fr)+(frL<=now-ed+fr);
ans=1LL*ans*kl%mod;
}
}
void work() {
for(RI i=1;i<=n;++i) {
if(du[i]) continue;
int x=i,len=0;while(!cir[x]) x=a[x],++len;
footL[x]=len;//算挂在每个点上的脚长
}
ans=1;
for(RI i=1;i<=n;++i) if(cir[i]) workcir(i);
for(RI i=1;i<=n;++i) {//对每一种长度的简单环做DP
if(!sum[i]) continue;
f[0]=1;
for(RI j=1;j<=sum[i];++j) {
if(i>1&&(i&1)) f[j]=qm(f[j-1]+f[j-1]);//情况1,2
else f[j]=f[j-1];//情况1
if(j>1) f[j]=qm(f[j]+1LL*f[j-2]*(j-1)%mod*i%mod);//情况3
}
ans=1LL*ans*f[sum[i]]%mod;
}
}
int main()
{
n=read();
for(RI i=1;i<=n;++i) a[i]=read(),++du[a[i]];
for(RI i=1;i<=n;++i) {
if(vis[i]) continue;
int x=i;while(!vis[x]) vis[x]=i,x=a[x];
if(vis[x]!=i) continue;//说明i在一个脚上
while(!cir[x]) cir[x]=1,x=a[x];//给环打上是环标记
}
for(RI i=1;i<=n;++i)//判无解
if((cir[i]&&du[i]>2)||(!cir[i]&&du[i]>1)) {puts("0");return 0;}
work();
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=5e5+500,mod=1000000007;
int p[N],stack[N],inq[N],vis[N],inring[N],len[N],tmp[N],cnt[N],top,n,rd[N],fac[N],ifac[N],inv[N],pw2[N],ipw2[N];
void dfs(int u)
{
stack[++top]=u,inq[u]=1,vis[u]=1;
int v=p[u];
if(inq[v])
{
int t=top;stack[top+1]=0;
while(stack[top+1]!=v)inring[stack[top--]]=1;
top=t;
}
if(!vis[v])dfs(v);inq[u]=0;
}
void getdep(int u)
{
int dep=0;while(!inring[u])++dep,u=p[u];
len[u]=dep;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",p+i),rd[p[i]]++;
for(int i=1;i<=n;i++)if(!vis[i])dfs(i);
for(int i=1;i<=n;i++)if(rd[i]>=2+inring[i]){puts("0");return 0;}
for(int i=1;i<=n;i++)if(!rd[i])getdep(i);
int ans=1;
for(int i=1;i<=n;i++)
{
if(!inring[i])continue;
int u=i,tn=0,iscir=!len[u];inring[u]=0;tmp[++tn]=u;
while(p[u]!=i)u=p[u],tmp[++tn]=u,iscir&=!len[u],inring[u]=0;
if(iscir){++cnt[tn];continue;}
int lst=0;
for(int j=tn;j>=1;j--)if(len[tmp[j]]){lst=j-tn;break;}
for(int j=1;j<=tn;j++)
{
if(!len[tmp[j]])continue;
if(j-len[tmp[j]]<lst){puts("0");return 0;}
if(j-len[tmp[j]]>lst)ans=(ans<<1)%mod;
lst=j;
}
}
fac[0]=1;for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
inv[1]=1;for(int i=2;i<=n;i++)inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
ifac[0]=1;for(int i=1;i<=n;i++)ifac[i]=1ll*ifac[i-1]*inv[i]%mod;
pw2[0]=1;for(int i=1;i<=n;i++)pw2[i]=(pw2[i-1]<<1)%mod;
ipw2[0]=1;for(int i=1;i<=n;i++)ipw2[i]=(ipw2[i-1]&1)?((ipw2[i-1]+mod)>>1):(ipw2[i-1]>>1);
for(int i=1;i<=n;i++)
{
if(!cnt[i])continue;
int tn=cnt[i];int s=0,pw=1;
for(int j=0;j*2<=tn;j++)
{
int t=1ll*ifac[tn-2*j]*ifac[j]%mod*ipw2[j]%mod*pw%mod;
if((i&1)&&i!=1)t=1ll*t*pw2[tn-2*j]%mod;
s=(s+t)%mod;pw=1ll*pw*i%mod;
}
ans=1ll*ans*s%mod*fac[tn]%mod;
}
cout<<ans<<endl;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }
const int MAXN = 100005, MOD = 1e9 + 7;
int n, arr[MAXN], vis[MAXN], cir[MAXN], inc[MAXN];
int dep[MAXN], cnt[MAXN];
LL fac[MAXN], inv[MAXN];
vector<int> gra[MAXN];
LL modpow(LL a, int b) {
LL res = 1;
for (; b; b >>= 1) {
if (b & 1) res = res * a % MOD;
a = a * a % MOD;
}
return res;
}
int dfs(int u) {
int cnt = 0, d = 0;
for (int v : gra[u]) {
if (inc[v]) continue;
if (cnt++) {
puts("0");
exit(0);
}
d = dfs(v) + 1;
}
return d;
}
int main() {
scanf("%d", &n);
fac[0] = 1;
for (int i = 1; i <= n; i++) {
scanf("%d", arr + i);
gra[arr[i]].push_back(i);
fac[i] = fac[i - 1] * i % MOD;
}
inv[n] = modpow(fac[n], MOD - 2);
for (int i = n; i > 0; i--) inv[i - 1] = inv[i] * i % MOD;
LL ans = 1;
for (int i = 1; i <= n; i++) if (!vis[i]) {
int j = i, tot = 0;
for (; !vis[j]; j = arr[j]) vis[j] = i;
if (vis[j] != i) continue;
for (int k = arr[j]; k != j; k = arr[k])
cir[tot++] = k, inc[k] = 1;
cir[tot++] = j, inc[j] = 1;
memset(dep, 0, sizeof(int) * tot);
bool flag = false;
for (int k = 0; k < tot; k++) {
int t = cir[k];
if (gra[t].size() == 1) continue;
dep[k] = dfs(t);
flag = true;
}
if (flag) {
for (int k = 0; k < tot; k++) if (dep[k]) {
int nxt = 0;
for (int j = (k - 1 + tot) % tot;; j = (j - 1 + tot) % tot) {
if (dep[j]) break;
++nxt;
}
if (nxt < dep[k] - 1) {
puts("0");
return 0;
}
if (nxt >= dep[k]) ans = ans * 2 % MOD;
}
} else ++cnt[tot];
}
for (int i = 1; i <= n; i++) if (cnt[i]) {
LL pw = 1, t = (LL)i * (MOD + 1) / 2 % MOD, sum = 0;
if (i & 1 && i > 1) pw = modpow(2, cnt[i]), t = (LL)i * (MOD + 1) / 8 % MOD;
for (int j = 0; 2 * j <= cnt[i]; j++) {
sum = (sum + pw * inv[cnt[i] - 2 * j] % MOD * inv[j]) % MOD;
pw = pw * t % MOD;
}
ans = ans * sum % MOD * fac[cnt[i]] % MOD;
}
printf("%lld\n", ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
typedef long long i64;
const int N=100007,P=1e9+7;
int n,fa[N],in[N],q[N],ql=0,qr=0,md[N],ss[N][2],t[N];
i64 pw2[N],fac[N],fiv[N],ans=1;
i64 pw(i64 a,i64 n){
i64 v=1;
for(;n;n>>=1,a=a*a%P)if(n&1)v=v*a%P;
return v;
}
int main(){
scanf("%d",&n);
for(int i=pw2[0]=1;i<=n;++i)pw2[i]=pw2[i-1]*2%P;
for(int i=fac[0]=1;i<=n;++i)fac[i]=fac[i-1]*i%P;
fiv[n]=pw(fac[n],P-2);
for(int i=n;i;--i)fiv[i-1]=fiv[i]*i%P;
for(int i=1;i<=n;++i)scanf("%d",fa+i),++in[fa[i]];
for(int i=1;i<=n;++i)if(!in[i])q[++qr]=i;
while(ql!=qr){
int w=q[++ql],f=fa[w];
if(md[f])return puts("0"),0;
md[f]=md[w]+1;
if(!--in[f])q[++qr]=f;
}
for(int i=1;i<=n;++i)if(in[i]){
int c=0,sp=0;
for(int w=i;in[w];in[w]=0,w=fa[w]){
++c;
if(md[w])ss[++sp][0]=md[w],ss[sp][1]=c;
}
if(sp){
ss[0][1]=ss[sp][1]-c;
for(int j=1;j<=sp;++j){
int d=ss[j][1]-ss[j-1][1],d0=ss[j][0];
if(d<d0)return puts("0"),0;
if(d>d0)ans=ans*2%P;
}
}else ++t[c];
}
for(int i=1;i<=n;++i)if(t[i]){
i64 s1=1,s2=0;
if(i>1&&(i&1))
for(int j=0,c=t[i];j<=c;j+=2){
s2=(s2+s1*pw2[c-j]%P*fiv[j/2])%P;
s1=i64(c-j)*(c-j-1)/2%P*s1%P*i%P;
}
else
for(int j=0,c=t[i];j<=c;j+=2){
s2=(s2+s1*fiv[j/2])%P;
s1=i64(c-j)*(c-j-1)/2%P*s1%P*i%P;
}
ans=ans*s2%P;
}
printf("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int sz=1e5+7;
const int mod=1e9+7;
int n;
int T;
int ans;
int cnt;
bool flag;
int b[sz];
int a[sz],r[sz],rr[sz];
int t[sz],rt[sz];
int qp[sz],_qp[sz];
int inv[sz],fac[sz],_fac[sz];
int del[sz],dep[sz];
queue<int>q;
void upd(int &x,int y){
x=x+y>=mod?x+y-mod:x+y;
}
int qpow(int x,int y){
int ret=1;
for(;y;y>>=1,x=1ll*x*x%mod) if(y&1) ret=1ll*x*ret%mod;
return ret;
}
void find(int x){
if(dep[x]) flag=0;
cnt++;
del[x]=1;
if(!del[a[x]]) find(a[x]);
b[a[x]]=x;
}
void init(){
qp[0]=_qp[0]=1;
fac[0]=_fac[0]=1;
inv[1]=fac[1]=_fac[1]=1;
for(int i=2;i<sz;i++){
inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
fac[i]=1ll*i*fac[i-1]%mod;
_fac[i]=1ll*inv[i]*_fac[i-1]%mod;
}
for(int i=1;i<sz;i++){
qp[i]=2ll*qp[i-1]%mod;
_qp[i]=1ll*inv[2]*_qp[i-1]%mod;
}
}
void build(){
for(int i=1;i<=n;i++) if(!r[i]) q.push(i);
while(!q.empty()){
int x=q.front();
dep[a[x]]=dep[x]+1;
r[a[x]]--;
if(!r[a[x]]) q.push(a[x]);
del[x]=1;
q.pop();
}
for(int i=1;i<=n;i++){
if(del[i]==1&&rr[i]>1) { printf("0\n"); exit(0); }
if(del[i]==0&&rr[i]>2) { printf("0\n"); exit(0); }
}
for(int i=1;i<=n;i++)
if(!del[i]){
cnt=0;
flag=1;
find(i);
if(flag) t[cnt]++;
else{
int x=i;
while(!dep[x]) x=b[x];
rt[++T]=x;
}
}
}
void dfs(int x,int res){
if(dep[x]){
if(res>0) ans=0;
if(res<0) ans=2ll*ans%mod;
res=dep[x];
}
if(!del[x]) return;
del[x]=0;
dfs(b[x],res-1);
}
int main(){
init();
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),r[a[i]]++,rr[a[i]]++;
build();
ans=1;
for(int i=1;i<=n;i++){
if(!t[i]) continue;
int sum=0;
for(int j=0;j<=t[i];j+=2){
int tot;
tot=1ll*fac[t[i]]*_fac[t[i]-j]%mod*_fac[j/2]%mod*_qp[j/2]%mod*qpow(i,j/2)%mod;
if(i%2==1&&i>1) tot=1ll*tot*qp[t[i]-j]%mod;
upd(sum,tot);
}
ans=1ll*ans*sum%mod;
}
for(int i=1;i<=T;i++) dfs(rt[i],0);
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define N 100000 + 5
#define Mod 1000000007
#define rep(i, l, r) for(int i = l; i <= r; ++i)
bool cir[N];
int n, ans, a[N], f[N], p[N], in[N], deg[N], vis[N], sum[N], inv[N], invp[N];
int read(){
char c; int x = 0, f = 1;
c = getchar();
while(c > '9' || c < '0'){ if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int Inc(int a, int b){ return (a += b) >= Mod ? a - Mod : a;}
int Mul(int a, int b){ return 1ll * a * b % Mod;}
int Qpow(int a, int b){
int ans = 1;
while(b){
if(b & 1) ans = Mul(ans, a);
a = Mul(a, a), b >>= 1;
}
return ans;
}
int C(int n, int m){
if(m > n) return 0;
return Mul(f[n], Mul(inv[m], inv[n - m]));
}
void solve(int p){
int x = p, fi = 0, fl = 0, last = 0, llen = 0, now = 0, nlen = 0, cnt = 0;
while(cir[x]){
cir[x] = false, cnt++; if(!in[x]){ x = a[x]; continue;}
if(!last) last = fi = cnt, llen = fl = in[x];
else{
if(!now) now = cnt, nlen = in[x];
else last = now, llen = nlen, now = cnt, nlen = in[x];
if(cnt - nlen < last) ans = 0;
if(cnt - nlen > last) ans = Mul(ans, 2);
}
x = a[x];
}
if(!fi) sum[cnt]++;
else{
if(!now) now = fi;
int len = cnt - (now - fi);
if(len < fl) ans = 0;
if(len > fl) ans = Mul(ans, 2);
}
}
int main(){
n = read();
rep(i, 1, n) a[i] = read(), deg[a[i]]++;
f[0] = inv[0] = p[0] = invp[0] = 1;
rep(i, 1, n){
f[i] = Mul(f[i - 1], i), inv[i] = Qpow(f[i], Mod - 2);
p[i] = Mul(p[i - 1], 2), invp[i] = Qpow(p[i], Mod - 2);
}
rep(i, 1, n) if(!vis[i]){
int x = i;
while(!vis[x]) vis[x] = i, x = a[x];
if(vis[x] == i) while(!cir[x]) cir[x] = true, x = a[x];
}
rep(i, 1, n) if((cir[i] && deg[i] >= 3) || (!cir[i] && deg[i] >= 2)){ puts("0"); return 0;}
rep(i, 1, n) if(!deg[i]){
int x = i, cnt = 0;
while(!cir[x]) x = a[x], cnt++;
in[x] = cnt;
}
ans = 1;
rep(i, 1, n) if(cir[i]) solve(i);
rep(i, 1, n) if(sum[i]){
int res = 0;
rep(j, 0, sum[i] / 2){
int cnt = 0;
cnt = C(sum[i], 2 * j);
cnt = Mul(cnt, C(2 * j, j));
cnt = Mul(cnt, f[j]);
cnt = Mul(cnt, invp[j]);
cnt = Mul(cnt, Qpow(i, j));
if((i & 1) && i != 1) cnt = Mul(cnt, p[sum[i] - 2 * j]);
res = Inc(res, cnt);
}
ans = Mul(ans, res);
}
printf("%d", ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
const int mod=1e9+7;
using namespace std;
int n,du[100010],vis[100010],nxt[100010],L[100010],ans=1,cir_len[100010],f[100010];
bool cir[100010];
void work_cir(int u){
int now=0,pre=0,f_pos=0,f_L=0;
while(cir[u]){
cir[u]=false,now++;
if(L[u]){
if(!f_pos) f_pos=pre=now,f_L=L[u];
else{
int k=(L[u]<now-pre)+(L[u]<=now-pre);
ans=1ll*ans*k%mod,pre=now;
}
}
u=nxt[u];
}
if(!f_pos) ++cir_len[now];
else{
int k=(f_L<now-pre+f_pos)+(f_L<=now-pre+f_pos);
ans=1ll*ans*k%mod;
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&nxt[i]),++du[nxt[i]];
for(int i=1;i<=n;i++){
if(vis[i]) continue;
int now=i;
while(!vis[now]) vis[now]=i,now=nxt[now];
if(vis[now]!=i) continue;
while(!cir[now]) cir[now]=true,now=nxt[now];
}
for(int i=1;i<=n;i++){
if((cir[i]&&du[i]>2)||(!cir[i]&&du[i]>1)){
puts("0\n");
return 0;
}
}
for(int i=1;i<=n;i++){
if(du[i]) continue;
int now=i,len=0;
while(!cir[now]) ++len,now=nxt[now];
L[now]=len;
}
for(int i=1;i<=n;i++) if(cir[i]) work_cir(i);
for(int i=1;i<=n;i++){
if(!cir_len[i]) continue;
f[0]=1;
for(int j=1;j<=cir_len[i];j++){
if(i>1&&(i&1)) f[j]=(2ll*f[j-1])%mod;
else f[j]=f[j-1];
if(j>1) f[j]=(f[j]+(1ll*f[j-2]*(j-1)%mod)*i%mod)%mod;
}
ans=1ll*ans*f[cir_len[i]]%mod;
}
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ll long long
#define MOD 1000000007
using namespace std;
inline int read(){
int re=0,flag=1;char ch=getchar();
while(!isdigit(ch)){
if(ch=='-') flag=-1;
ch=getchar();
}
while(isdigit(ch)) re=(re<<1)+(re<<3)+ch-'0',ch=getchar();
return re*flag;
}
ll f[200010],finv[200010],meth[200010];
ll qpow(ll a,ll b){
ll re=1;
while(b){
if(b&1) re=(re*a)%MOD;
a=a*a%MOD;b>>=1;
}
return re;
}
void init(){
ll i,len=200000;
f[0]=f[1]=finv[0]=finv[1]=1;
for(i=2;i<=len;i++) f[i]=f[i-1]*i%MOD;
finv[len]=qpow(f[len],MOD-2);
for(i=len;i>2;i--) finv[i-1]=finv[i]*i%MOD;
}
int n,a[200010],vis[200010],cir[200010],cntcir=0,in[200010],bst[200010],siz[200010];
ll ans=1;
vector<int>s;
vector<int>nd[200010];
bool cmp(int l,int r){
return siz[l]<siz[r];
}
ll C(ll x,ll y){
return f[x]*finv[y]%MOD*finv[x-y]%MOD;
}
int main(){
n=read();int i,j;ll tmp,c,cc;
init();
meth[0]=1;
for(i=1;i<=n;i++) meth[i]=C(i*2,i)*f[i]%MOD*qpow(qpow(2,MOD-2),i)%MOD;
for(i=1;i<=n;i++) a[i]=read(),in[a[i]]++;
for(i=1;i<=n;i++){
j=i;
while(!vis[j]) vis[j]=i,j=a[j];
// cout<<"get "<<i<<' ' <<j<<'\n';
if(vis[j]^i) continue;
// cout<<"getnew "<<i<<' ' <<j<<'\n';
cntcir++;
while(!cir[j]){
// if(cntcir==20) cout<<"getcir "<<j<<' '<<a[j]<<' '<<cntcir<<'\n';
cir[j]=cntcir,nd[cntcir].push_back(j),siz[cntcir]++,j=a[j];
}
}
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++){
if(in[i]) continue;
j=i;
while(!cir[j]&&!vis[j]) j=a[j];
if(vis[j]){
puts("0");return 0;
}
// cout<<"find "<<i<<' '<<j<<'\n';
bst[cir[j]]=1;tmp=cir[j];
j=i;c=0;
while(!cir[j]) cir[j]=tmp,c++,vis[j]=1,j=a[j];
vis[j]=c;
// cout<<"finish add "<<cir[j]<<' '<<c<<'\n';
}
for(i=1;i<=cntcir;i++){
// cout<<"check link "<<i<<' '<<bst[i]<<'\n';
if(!bst[i]){s.push_back(i);continue;}
for(j=0;j<nd[i].size();j++) if(vis[nd[i][j]]) break;
tmp=j;
do{
c=j;cc=1;
j--;
(j+=(int)nd[i].size());
j%=(int)nd[i].size();
for(;!vis[nd[i][j]];j--,j=((j<0)?j+nd[i].size():j)) cc++;
// cout<<" end "<<c<<' '<<j<<' '<<cc<<' '<<vis[nd[i][c]]<<'\n';
if(vis[nd[i][c]]>cc){
puts("0");return 0;
}
// cout<<"survive "<<ans<<"\n";
if(vis[nd[i][c]]<cc) (ans*=2)%=MOD;
}while(tmp!=j);
}
sort(s.begin(),s.end(),cmp);
for(i=0;i<s.size();i+=c){
j=i;tmp=0;
while(siz[s[j]]==siz[s[i]]&&j<s.size()) j++;
// cout<<"circles "<<i<<' '<<j<<'\n';
c=j-i;
for(j=0;j<=c/2;j++){
(tmp+=C(c,2*j)*meth[j]%MOD*qpow(siz[s[i]],j)%MOD*qpow(2,(c-2*j)*(siz[s[i]]!=1)*(siz[s[i]]&1))%MOD)%=MOD;
// cout<<' '<<j<<' '<<tmp<<'\n';
}
ans=ans*tmp%MOD;
}
printf("%lld\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | // #include {{{
#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <bitset>
#include <vector>
#include <complex>
#include <algorithm>
using namespace std;
// }}}
// #define {{{
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define per(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
// }}}
const int N = 1e5 + 10 , P = 1e9 + 7;
int n , a[N] , du[N] , in[N] , sz[N];
int circlecnt[N] , f[N];
int main(){
cin >> n;
rep(i,1,n+1) cin >> a[i] , in[a[i]]++;
rep(i,1,n+1) du[i] = in[i] , sz[i] = 1;
vi q;
rep(i,1,n+1) if(!du[i]) q.pb(i);
rep(i,0,sz(q)) {
int c=q[i];
sz[a[c]]+=sz[c];
if(!--du[a[c]]) q.pb(a[c]);
}
rep(i,1,n+1) if(in[i]>2||(in[i]==2&&!du[i]))
return puts("0") , 0;
int ans = 1;
rep(i,1,n+1) if(du[i]) {
vi cir;
for(int c=i;du[c];c=a[c])
cir.pb(c),du[c]=0;
vi branch;
rep(i,0,sz(cir)) if(sz[cir[i]]!=1)
branch.pb(i);
if(sz(branch)) {
branch.pb(branch[0]);
rep(i,1,sz(branch)) {
int from = branch[i - 1] , to = branch[i];
int dis = to - from;
if(dis <= 0) dis += sz(cir);
if(dis < sz[cir[to]] - 1)
return puts("0") , 0;
else if(dis > sz[cir[to]] - 1)
(ans *= 2) %= P;
}
} else circlecnt[sz(cir)]++;
}
rep(i,1,n+1) if(circlecnt[i]) {
int x = circlecnt[i];
f[0] = 1;
rep(j,1,x+1) {
f[j] = f[j - 1];
if(i >= 3 && i % 2 == 1) (f[j] += f[j - 1]) %= P;
if(j >= 2) (f[j] += ll(f[j - 2]) * (j - 1) % P * i % P) %= P;
}
ans = ll(ans) * f[x] % P;
}
cout << ans << endl;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define f(a,b,c) for(int a=(b);a<=(c);++a)
const int mod=1e9+7,maxn=1e5+5;
int a[maxn],d[maxn],ci[maxn],vis[maxn],fl[maxn],sum[maxn],f[maxn];
int add(int a,int b){a+=b;return a>=mod?a-mod:a;}
int n,ans=1;
#define GG exit((puts("0"),0))
void solve(int x)//基环树
{
int cur,ths,lst,fst;
cur=ths=lst=fst=0;
while(ci[x])
{
++cur,ci[x]=0;
if(fl[x])
{
if(ths==0)ths=lst=cur,fst=fl[x];
else
{
if(fl[x]>cur-lst)GG;
if(fl[x]==cur-lst)ans=ans;
if(fl[x]<cur-lst)ans=add(ans,ans);
lst=cur;
}
}
x=a[x];
}
if(!ths)sum[cur]++;
else
{
if(fst>cur-lst+ths)GG;
if(fst==cur-lst+ths)ans=ans;
if(fst<cur-lst+ths)ans=add(ans,ans);
}
}
int main()
{
scanf("%d",&n);
f(i,1,n)scanf("%d",a+i),d[a[i]]++;
int x,l;
f(i,1,n)
{
if(!vis[i])
{
for(x=i;!vis[x];vis[x]=i,x=a[x]);
if(vis[x]!=i)continue;
while(!ci[x])ci[x]=1,x=a[x];
}
}
f(i,1,n){
if(ci[i]&&d[i]>=3)GG;
if(!ci[i]&&d[i]>=2)GG;
}
f(i,1,n)
{
if(d[i])continue;
for(x=i,l=0;!ci[x];x=a[x],l++);
fl[x]=l;
}
f(i,1,n)if(ci[i])solve(i);//基环树
f(i,1,n)//简单环
{
f[0]=1;
f(j,1,sum[i])
{
if(i>1&&i%2==1)f[j]=add(f[j-1],f[j-1]);
else f[j]=f[j-1];
if(j>1)f[j]=add(f[j],(ll)f[j-2]*(j-1)%mod*i%mod);
}
ans=(ll)ans*f[sum[i]]%mod;
}
printf("%d",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<cmath>
#include<string>
#define ls (t<<1)
#define rs ((t<<1)+1)
#define mid ((l+r)>>1)
#define fi first
#define se second
#define mk make_pair
#define pb push_back
#define N 100005
#define M 200005
#define Mo 1000000007
using namespace std;
int i,j,m,n,p,k,a[N],chun[N],deg[N],D[N],vis[N],Q[N],size[N],tot,ans,Qt[N];
int fac[N],inv[N];
vector<int>tree[N],v[N];
void work(int x)
{
int i;
Q[0]=0;
for (i=x;!vis[x];x=a[x]) vis[x]=1,Q[++Q[0]]=x;
for (i=1;i<=Q[0];++i) if (size[Q[i]]) break;
if (i<=Q[0])
{
++tot;
for (i=1;i<=Q[0];++i) tree[tot].pb(Q[i]);
}
else chun[Q[0]]++;
}
void jia(int &x,int y)
{
x+=y; if (x>=Mo) x-=Mo;
}
int dp()
{
int i,la=0,lb=0,fa=1;
for (i=1;i<=Qt[0];++i)
if (size[Qt[i]])
{
int na=i+size[Qt[i]]-1,nb=i+size[Qt[i]],ga=0;
if (la<i) jia(ga,fa);
if (lb&&lb<i) jia(ga,fa);
fa=ga; la=na; lb=nb;
}
int now=0;
if (la<=Qt[0]) jia(now,fa);
if (lb<=Qt[0]) jia(now,fa);
return now;
}
int power(int x,int y)
{
int sum=1;
for (;y;y>>=1)
{
if (y&1) sum=1ll*sum*x%Mo;
x=1ll*x*x%Mo;
}
return sum;
}
int C(int x,int y)
{
return 1ll*fac[x]*inv[y]%Mo*inv[x-y]%Mo;
}
void pre()
{
fac[0]=inv[0]=1;
for (i=1;i<N;++i) fac[i]=1ll*fac[i-1]*i%Mo,inv[i]=power(fac[i],Mo-2);
}
int Count(int x)
{
Q[0]=0;
int i;
for (i=0;i<(int)tree[x].size();++i) Q[++Q[0]]=tree[x][i];
if (Q[0]==1)
{
if (size[Q[1]]==1) return 1;
return 0;
}
for (i=1;i<=Q[0];++i) if (size[Q[i]]) break;
Qt[0]=0;
for (j=i;j>=1;--j) Qt[++Qt[0]]=Q[j];
for (j=Q[0];j>i;--j) Qt[++Qt[0]]=Q[j];
return dp();
}
int main()
{
pre();
scanf("%d",&n);
for (i=1;i<=n;++i) scanf("%d",&a[i]),v[i].pb(a[i]),deg[a[i]]++;
for (i=1;i<=n;++i) D[i]=deg[i];
for (i=1;i<=n;++i) if (!deg[i]) Q[++Q[0]]=i;
for (i=1;i<=n;++i) if (deg[i]>2)
{
puts("0");
return 0;
}
int l;
for (l=1;l<=Q[0];++l)
{
int p=Q[l]; vis[p]=1;
size[p]++;
for (i=0;i<(int)v[p].size();++i)
{
int k=v[p][i];
deg[k]--; size[k]+=size[p];
if (deg[k]==0) Q[++Q[0]]=k;
}
}
for (i=1;i<=n;++i) if (vis[i]&&D[i]==2)
{
puts("0");
return 0;
}
for (i=1;i<=n;++i) if (!vis[i]) work(i);
ans=1;
for (i=1;i<=n;++i)
if (chun[i])
{
int now=0;
for (j=0;j<=chun[i];j+=2)
jia(now,
1ll*((i>1&&(i&1))?power(2,chun[i]-j):1)*C(chun[i],j)%Mo*C(j,j/2)%Mo*fac[j/2]%Mo*power(power(2,j/2),Mo-2)%Mo*power(i,j/2)%Mo);
ans=1ll*ans*now%Mo;
}
for (i=1;i<=tot;++i)
{
int now=Count(i);
ans=1ll*ans*now%Mo;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#define MN 100000
#define mod 1000000007
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int head[MN+5],cnt=0,n,vis[MN+5],Fa[MN+5],dep[MN+5],c[MN+5],top,in[MN+5],ans=1;
int A,B,C,num[MN+5],vnum,cirnum[MN+5],p[MN+5],inv[MN+5],pw[MN+5],Out[MN+5];
struct edge{int to,next;}e[MN*2+5];
inline void ins(int f,int t)
{
e[++cnt]=(edge){t,head[f]};head[f]=cnt;
e[++cnt]=(edge){f,head[t]};head[t]=cnt;
}
void dfs(int x,int fa)
{
Fa[x]=fa;vis[x]=1;++vnum;
for(int i=head[x];i;i=e[i].next)
if(e[i].to!=fa)
{
if(vis[e[i].to]) A=x,B=e[i].to;
else dep[e[i].to]=dep[x]+1,dfs(e[i].to,x);
}
}
inline int lca(int x,int y)
{
for(;x!=y;x=Fa[x])
if(dep[x]<dep[y]) swap(x,y);
return x;
}
int GetChain(int x,int fa)
{
int son=0,val=1;
for(int i=head[x];i;i=e[i].next)
if(e[i].to!=fa)
{
++son;
int res=GetChain(e[i].to,x);
if(res==0) return 0;
else val+=res;
}
return son>1?0:val;
}
inline int CC(int n,int m){return 1LL*p[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
n=read();pw[0]=p[0]=inv[0]=inv[1]=p[1]=1;pw[1]=2;
for(int i=2;i<=n;++i) p[i]=1LL*p[i-1]*i%mod,inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod,pw[i]=2*pw[i-1]%mod;
for(int i=2;i<=n;++i) inv[i]=1LL*inv[i-1]*inv[i]%mod;
for(int i=1,j;i<=n;++i) ++num[j=read()],ins(i,j);
for(int i=1;i<=n;++i) if(num[i]>2) return 0*puts("0");
for(int i=1;i<=n;++i) if(!vis[i])
{
vnum=top=0;dfs(i,0);C=lca(A,B);
int len=dep[A]+dep[B]-2*dep[C]+1;
if(len==vnum) {++cirnum[vnum];continue;}
for(;A!=C;A=Fa[A]) c[++top]=A;c[++top]=C;top=len;
for(;B!=C;B=Fa[B]) c[len--]=B;
for(int j=1;j<=top;++j) in[c[j]]=1;
for(int j=head[c[1]];j;j=e[j].next)
if(e[j].to==c[2])
{
if(~j&1)
for(int l=1,r=top;l<r;++l,--r)
swap(c[l],c[r]);
break;
}
for(int j=1;j<=top;++j)
{
Out[j]=0;
for(int k=head[c[j]];k;k=e[k].next)
if(!in[e[k].to])
{
Out[j]=GetChain(e[k].to,c[j]);
if(!Out[j]) return 0*puts("0");
}
}
int last,way=1;
for(int j=top;j;--j) if(Out[j]) {last=top-j;break;}
for(int j=1;j<=top;++j)
{
++last;
if(!Out[j]) continue;
if(Out[j]>last) return 0*puts("0");
else if(Out[j]<last) way=way*2%mod;
last=0;
}
ans=1LL*ans*way%mod;
}
for(int i=1;i<=n;++i) if(cirnum[i])
{
int way=(i==1||i%2==0)?1:pw[cirnum[i]];
for(int j=1,k=i,i2=mod+1>>1;j<<1<=cirnum[i];++j,k=1LL*k*i%mod,i2=1LL*i2*(mod+1)/2%mod)
{
int thway=1LL*CC(cirnum[i],j*2)*CC(j*2,j)%mod*p[j]%mod*i2%mod*k%mod;
if(i==1||i%2==0);else thway=1LL*thway*pw[cirnum[i]-2*j]%mod;
way=(way+thway)%mod;
}
ans=1LL*ans*way%mod;
}
cout<<ans;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int N=100010,mo=int(1e9+7);
struct edge{int s,t,n;}e[N];
int n,a[N],h[N],d[N],vis[N],inc[N],len[N],cnt[N],fac[N],ifac[N],ans;
vector <int> V;
int qpow(int a,int b)
{
int x=a; a=1;
while (b)
{
if (b&1) a=1LL*a*x%mo;
x=1LL*x*x%mo,b>>=1;
}
return a;
}
int C(int n,int m){return 1LL*fac[n]*ifac[m]%mo*ifac[n-m]%mo;}
int go(int x)
{
if (vis[x]) return x;
vis[x]=1;
int _=go(a[x]);
if (_) inc[x]=1,V.push_back(x);
return _==x?0:_;
}
int dfs(int x,int f)
{
int _=0;
vis[x]=1;
for (int i=h[x],y; y=e[i].t,i; i=e[i].n)
if ((y!=f)&&(!inc[y]))
{
if (_) return -1<<30;
else _=dfs(y,x)+1;
}
return _;
}
void work()
{
scanf("%d",&n),ans=1;
for (int i=1; i<=n; i++)
scanf("%d",&a[i]),e[i]=(edge){a[i],i,h[a[i]]},h[a[i]]=i,d[a[i]]++;
for (int i=1,s; i<=n; i++)
if (!vis[i])
{
V.clear(),go(i);
for (int i=0; i<V.size(); i++)
if ((len[i+1]=dfs(V[i],0))<0)
puts("0"),exit(0);
s=0;
for (int i=1; i<=V.size(); i++)
if (!len[i]) s++; else break;
if (s==V.size()) {cnt[s]++; continue;}
for (int i=V.size(); i; i--)
{
s++;
if (len[i])
{
if (s<len[i]) puts("0"),exit(0);
if (s>len[i]) ans=2*ans%mo;
s=0;
}
}
}
fac[0]=1;
for (int i=1; i<=n; i++) fac[i]=1LL*fac[i-1]*i%mo;
ifac[n]=qpow(fac[n],mo-2);
for (int i=n; i; i--) ifac[i-1]=1LL*ifac[i]*i%mo;
for (int i=1,s; i<=n; i++)
if (cnt[i])
{
s=0;
for (int j=0,_; j<=cnt[i]/2; j++)
{
_=1LL*C(cnt[i],2*j)*fac[2*j]%mo*ifac[j]%mo*qpow(2,mo-1-j)%mo*qpow(i,j)%mo;
if ((i&1)&&(i>1)) _=1LL*_*qpow(2,cnt[i]-2*j)%mo;
s=(s+_)%mo;
}
ans=1LL*ans*s%mo;
}
printf("%d",ans);
}
int main()
{
work();
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int RLEN=1<<18|1;
inline char nc() {
static char ibuf[RLEN],*ib,*ob;
(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
char ch=nc(); int i=0,f=1;
while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
return i*f;
}
const int N=1e5+50, mod=1e9+7;
inline int add(int x,int y) {return (x+y>=mod) ? (x+y-mod) : (x+y);}
inline int dec(int x,int y) {return (x-y<0) ? (x-y+mod) : (x-y);}
inline int mul(int x,int y) {return (long long)x*y%mod;}
inline int power(int a,int b,int rs=1) {for(;b;b>>=1,a=mul(a,a)) if(b&1) rs=mul(rs,a); return rs;}
inline int cinv(int a) {return power(a,mod-2);}
int n,ans=1,fa[N],incir[N],vis[N];
int cnt[N];
vector <int> edge[N];
struct combin {
int fac[N],ifac[N];
combin() {
fac[0]=1;
for(int i=1;i<N;i++) fac[i]=mul(fac[i-1],i);
ifac[N-1]=cinv(fac[N-1]);
for(int i=N-2;~i;i--) ifac[i]=mul(ifac[i+1],i+1);
}
inline int C(int a,int b) {return (a<b) ? 0 : (mul(fac[a],mul(ifac[b],ifac[a-b])));}
} C;
inline void dfs(int x) {
vis[x]=1;
for(auto v:edge[x]) {
if(!vis[v]) fa[v]=x, dfs(v);
else {
for(int u=x;;u=fa[u]) {
if(!incir[u]) incir[u]=1;
else {puts("0"); exit(0);}
if(u==v) break;
}
}
}
}
inline int calc_dep(int x) {
if(edge[x].size()>=2) return -1e9;
if(!edge[x].size()) return 1;
return calc_dep(edge[x][0])+1;
}
vector <int> cir;
vector <int> ft;
int pos[N],dep[N],f[N];
inline int calc(int x) {
cir.clear(); ft.clear();
int now=x;
while(1) {
cir.push_back(now); pos[now]=cir.size();
vis[now]=1;
if(edge[now].size()>=3) return 0;
for(auto v:edge[now]) {
if(!incir[v]) {
int t=calc_dep(v);
if(t>0) ft.push_back(now), dep[now]=t;
else return 0;
}
}
for(auto v:edge[now]) if(incir[v]) {now=v; break;}
if(now==x) break;
}
if(!ft.size()) return ++cnt[cir.size()],1;
else {
int sum=1;
for(int i=0;i<ft.size();++i) {
int dis=(ft.size()==1) ? cir.size() : (ft[i]==ft.back() ? cir.size()-pos[ft[i]]+pos[ft[0]] : pos[ft[i+1]]-pos[ft[i]]);
if(dis<dep[ft[i]]) return 0;
else if(dis>dep[ft[i]]) sum=mul(sum,2);
} return sum;
}
}
int main() {
n=rd();
for(int i=1;i<=n;i++)
edge[rd()].push_back(i);
for(int i=1;i<=n;i++)
if(!vis[i] && edge[i].size()>=2) dfs(i);
for(int i=1;i<=n;i++)
if(!vis[i]) dfs(i);
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
if(incir[i] && !vis[i] && ans) ans=mul(ans,calc(i));
f[0]=f[2]=1;
for(int i=4;i<=n;i+=2) f[i]=add(f[i-2],mul(mul(i-2,i-3),f[i-4]));
for(int i=1;i<=n;i++) {
int sum=0;
for(int j=0;j<=cnt[i];j+=2) {
int tp=mul(C.C(cnt[i],j),f[j]);
if(i!=1) {
if(i&1) tp=mul(tp,power(2,cnt[i]-j));
tp=mul(tp,power(i,j/2));
} sum=add(sum,tp);
} ans=mul(ans,sum);
}
cout<<ans<<'\n';
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
int q = 0;
char ch = ' ';
while (ch < '0' || ch > '9') ch = getchar();
while (ch >= '0' && ch <= '9') q = q * 10 + ch - '0', ch = getchar();
return q;
}
const int mod = 1e9 + 7, N = 100005;
int n, ans;
int a[N], du[N], cir[N], vis[N], footL[N], sum[N], f[N];
int qm(int x) { return x >= mod ? x - mod : x; }
void workcir(int x) {
int now = 0, fr = 0, ed = 0, frL = 0;
while (cir[x]) {
++now, cir[x] = 0;
if (footL[x]) {
if (!fr)
ed = fr = now, frL = footL[x];
else {
int kl = (footL[x] < now - ed) + (footL[x] <= now - ed);
ans = 1LL * ans * kl % mod, ed = now;
}
}
x = a[x];
}
if (!fr)
++sum[now];
else {
int kl = (frL < now - ed + fr) + (frL <= now - ed + fr);
ans = 1LL * ans * kl % mod;
}
}
void work() {
for (RI i = 1; i <= n; ++i) {
if (du[i]) continue;
int x = i, len = 0;
while (!cir[x]) x = a[x], ++len;
footL[x] = len;
}
ans = 1;
for (RI i = 1; i <= n; ++i)
if (cir[i]) workcir(i);
for (RI i = 1; i <= n; ++i) {
if (!sum[i]) continue;
f[0] = 1;
for (RI j = 1; j <= sum[i]; ++j) {
if (i > 1 && (i & 1))
f[j] = qm(f[j - 1] + f[j - 1]);
else
f[j] = f[j - 1];
if (j > 1) f[j] = qm(f[j] + 1LL * f[j - 2] * (j - 1) % mod * i % mod);
}
ans = 1LL * ans * f[sum[i]] % mod;
}
}
int main() {
n = read();
for (RI i = 1; i <= n; ++i) a[i] = read(), ++du[a[i]];
for (RI i = 1; i <= n; ++i) {
if (vis[i]) continue;
int x = i;
while (!vis[x]) vis[x] = i, x = a[x];
if (vis[x] != i) continue;
while (!cir[x]) cir[x] = 1, x = a[x];
}
for (RI i = 1; i <= n; ++i)
if ((cir[i] && du[i] > 2) || (!cir[i] && du[i] > 1)) return puts("0"), 0;
work();
printf("%d\n", ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int N=100005,P=1e9+7;
typedef long long ll;
int n,ans=1,crc[N],nx[N],f0,f1,f2;
vector<int>e[N];
bool vis[N],in[N];
inline void Assert(bool f){if(!f)puts("0"),exit(0);}
int find(int x){if(in[x])return x;in[x]=true;int R=find(nx[x]);return in[x]=false,R;}
int dfs(int x){vis[x]=true;Assert(e[x].size()<=1);for(int v:e[x])return dfs(v)+1;return 1;}
void check(int x){
int u=find(x),m=1;
static vector<int>cr,le;
cr.clear();cr.pb(u);in[u]=vis[u]=true;
for(int p=nx[u];p!=u;p=nx[p])cr.pb(p),in[p]=vis[p]=true,m++;
le.resize(cr.size());
int f=-1;
for(int i=0,v;i<m;i++){
Assert(e[v=cr[i]].size()<=2);
for(int p:e[v])if(!in[p])le[i]=dfs(p),f=i;
if(e[v].size()==1)le[i]=0;
}
if(f==-1){crc[m]++;return;}
for(int i=0;i<m;i++){
if(le[i]){
int dis=(i-f+m)%m;
if(f==i)dis=m;
Assert(dis>=le[i]);
if(dis>le[i])ans=(ans<<1)%P;
f=i;
}
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&nx[i]),e[nx[i]].pb(i);
for(int i=1;i<=n;i++)if(!vis[i])check(i);
for(int i=1;i<=n;i++)if(crc[i]){
int c0=1+(i%2==1&&i>=3);
f0=1,f1=c0;
for(int j=2;j<=crc[i];j++){
f2=((ll)(j*i-i)*f0%P+c0*f1)%P;
f0=f1,f1=f2;
}
ans=(ll)ans*f1%P;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 200010
#define LL long long
using namespace std;
const LL P=1000000007;
LL getPow(LL x,LL y){
LL res=1;
while(y){
if(y&1) res=res*x%P;
x=x*x%P;
y>>=1;
}
return res;
}
struct edge{
int to,next;
edge(int _to=0,int _next=0):to(_to),next(_next){}
}e[MAXN];
int n;
int g[MAXN],nume;
int p[MAXN];
bool visit0[MAXN],visit[MAXN];
int b[MAXN],l[MAXN],sl[MAXN],numb;
int cnt[MAXN];
LL f[MAXN],fac[MAXN],invfac[MAXN];
LL ans=1;
void addEdge(int u,int v){
e[nume]=edge(v,g[u]);
g[u]=nume++;
}
int dfs(int x){
visit[x]=1;
int numch=0;
int res=0;
for(int i=g[x];~i;i=e[i].next)
if(!visit[e[i].to]){
numch++;
res=dfs(e[i].to)+1;
}
if(numch>=2) ans=0;
return res;
}
void gao(int x){
numb=0;
while(!visit[x]){
visit[x]=1;
b[++numb]=x;
x=p[x];
}
int pos=0;
for(int i=1;i<=numb;i++){
l[i]=dfs(b[i]);
if(l[i]) pos=i;
}
if(!pos){
cnt[numb]++;
return;
}
for(int i=1;i<=numb;i++){
b[i+numb]=b[i];
l[i+numb]=l[i];
b[i]=b[i+pos-1];
l[i]=l[i+pos-1];
}
for(int i=1;i<=numb;i++) sl[i]=sl[i-1]+l[i];
LL res1=0,res2=0;
if(sl[numb]-sl[numb-(l[1]-1)]==0){
res1=1;
for(int i=2;i<=numb;i++){
if(!l[i]) continue;
LL temp=0;
if(sl[i-1]-sl[max(0,i-l[i])]==0) temp++;
if(sl[i-1]-sl[max(0,i-l[i]-1)]==0) temp++;
res1=res1*temp%P;
}
}
if(sl[numb]-sl[numb-l[1]]==0){
res2=1;
for(int i=2;i<=numb;i++){
if(!l[i]) continue;
LL temp=0;
if(sl[i-1]-sl[max(0,i-l[i])]==0) temp++;
if(sl[i-1]-sl[max(0,i-l[i]-1)]==0) temp++;
res2=res2*temp%P;
}
}
ans=ans*(res1+res2)%P;
}
void init(){
f[0]=1;
for(int i=2;i<=n;i++) f[i]=f[i-2]*(i-1)%P;
fac[0]=1;
for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;
invfac[n]=getPow(fac[n],P-2);
for(int i=n-1;i>=0;i--) invfac[i]=invfac[i+1]*(i+1)%P;
}
LL getC(int x,int y){
return fac[x]*invfac[y]%P*invfac[x-y]%P;
}
int main(){
#ifdef DEBUG
freopen("E.in","r",stdin);
#endif
memset(g,-1,sizeof g);
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",p+i);
addEdge(p[i],i);
}
for(int i=1;i<=n;i++)
if(!visit[i]){
int x=i;
for(;!visit0[x];x=p[x]) visit0[x]=1;
gao(x);
}
init();
for(int i=1;i<=n;i++)
if(cnt[i]){
LL res=0;
for(int j=0;j<=cnt[i];j+=2)
if(i>=3 && (i&1)) res=(res+getC(cnt[i],j)*f[j]%P*getPow(i,j/2)%P*getPow(2,cnt[i]-j))%P;
else res=(res+getC(cnt[i],j)*f[j]%P*getPow(i,j/2))%P;
ans=ans*res%P%P;
}
printf("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int read(){ int x=0; char ch=getchar(); while (!isdigit(ch)) ch=getchar(); while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar(); return x; }
const int N=100005,mod=1e9+7;
int n,a[N],in[N],size[N],vis[N],cro[N]; int q[N],head,tail; int k[N],m; int r[N]; void add(int &x,int y){ if ((x+=y)>=mod) x-=mod; } int sc=0;
int Pre(int k,int n){ return k==1?n:(k-1); }
int solve(int n){
int ans=1,f=0;
for (int i=1;i<=n;i++){
if (k[i]==0) continue;
f=1; int x=Pre(i,n); while (!k[x]) x=Pre(x,n); int t=(i-x+n)%n;
if (t==0) t+=n; if (t<k[i]) return 0; if (t>k[i]) ans=2LL*ans%mod;
}
if (!f) r[n]++;
return ans;
}
int dp[N];
int main(){
n=read();
memset(in,0,sizeof in);
for (int i=1;i<=n;i++){
in[a[i]=read()]++;
size[i]=1;
}
head=tail=0;
for (int i=1;i<=n;i++){
if (in[i]>2)
return puts("0"),0;
if (in[i]==0)
q[++tail]=i;
if (in[i]==2)
cro[i]=1;
}
while (head!=tail){
int x=q[++head],y=a[x];
size[y]+=size[x];
if (!(--in[y]))
q[++tail]=y;
}
for (int i=1;i<=n;i++)
if (cro[i]&&!in[i])
return puts("0"),0;
int ans=1;
memset(vis,0,sizeof vis);
for (int i=1;i<=n;i++){
if (in[i]==0||vis[i])
continue;
m=0;
for (int x=i;!vis[x];x=a[x])
vis[x]=1,k[++m]=size[x]-1;
ans=1LL*ans*solve(m)%mod;
}
for (int i=1;i<=n;i++)
if (r[i]>0){
dp[0]=1,dp[1]=i>1&&(i&1)?2:1;
for (int j=2;j<=r[i];j++)
dp[j]=(1LL*(j-1)*i%mod*dp[j-2]+dp[j-1]*(i>1&&(i&1)?2:1))%mod;
ans=1LL*ans*dp[r[i]]%mod;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;
int p[N],vis[N],st[N],len[N],tot[N];
int head[N],nxt[N];
bool rt[N];
int f[N],g[N];
inline int dfs(int k) {
int s=0;
for (int i=head[k];i;i=nxt[i])
if (!rt[i]) {
if (s) puts("0"),exit(0);
s=dfs(i)+1;
}
return s;
}
int main()
{
const int mod=1e9+7;
int n,i,k,t,s,cnt=0,top,ans=1,p2=0;
bool is;
for (i=1,cin>>n;i<=n;i++) cin>>p[i],nxt[i]=head[p[i]],head[p[i]]=i;
for (i=1;i<=n;i++)
if (!vis[i]) {
++cnt;
for (k=i;!vis[k];k=p[k])
vis[k]=cnt;
if (vis[k]!=cnt) continue;
for (t=k,top=0;st[1]!=t;t=p[t]) rt[st[++top]=t]=true;
reverse(st+1,st+1+top);
is=false;
for (k=1;k<=top;k++)
is|=len[k]=dfs(st[k]);
if (is) {
for (k=1;k<=top;k++)
if (len[k]) {
for (t=k%top+1,s=1;!len[t];t=t%top+1,s++);
if (s<len[k]) return puts("0"),0;
if (s!=len[k]) p2++;
}
} else tot[top]++,p2+=top>1&&(top&1);
}
const int inv4=1LL*(mod+1)*(mod+1)/4%mod;
for (i=1;i<=n;i++) {
for (k=f[0]=f[1]=g[0]=g[1]=1;k<=tot[i];k++) {
f[k+1]=(1LL*f[k-1]*k%mod*inv4%mod*i+f[k])%mod;
g[k+1]=(1LL*g[k-1]*k%mod*i+g[k])%mod;
}
ans=1LL*ans*(i==1||i%2==0?g[tot[i]]:f[tot[i]])%mod;
}
while (p2--) (ans<<=1)%=mod;
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define N 100005
#define ll long long
#define P 1000000007
using namespace std;
int n,fa[N],b[N],rd[N],fl[N],sum[N];
void biu()
{
puts("0");
exit(0);
}
void dfs(int x)
{
if (fl[fa[x]]) biu();
rd[fa[x]]--;fl[fa[x]]=fl[x]+1;rd[x]=-1;
if (rd[fa[x]]==0) dfs(fa[x]);
}
int F(int x,int k)
{
static int f[N];
f[0]=1;f[1]=2;
for (int i=2;i<=x;i++)
f[i]=(f[i-1]*2+(ll)f[i-2]*(i-1)%P*k)%P;
return f[x];
}
int G(int x,int k)
{
// cout<<x<<' '<<k<<endl;
static int g[N];
g[0]=1;g[1]=1;
for (int i=2;i<=x;i++)
g[i]=(g[i-1]+(ll)g[i-2]*(i-1)%P*k)%P;
return g[x];
}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
scanf("%d",&fa[i]),rd[fa[i]]++;
for (int i=1;i<=n;i++)
if (rd[i]==0) dfs(i);
int Ans=1;
for (int i=1;i<=n;i++)
if (rd[i]==1)
{
int x=i,y=fa[i],cnt=0,flag=0;
b[++cnt]=fl[i];
if (fl[i]) flag=cnt;
for (;y!=i;y=fa[y],x=fa[x])
{
b[++cnt]=fl[y];rd[y]=0;
if (fl[y]) flag=cnt;
}
//cout<<cnt<<' '<<flag<<endl;
if (!flag) sum[cnt]++;
else
{
flag=flag-cnt;
for (int j=1;j<=cnt;j++)
{
//cout<<b[j]<<endl;
if (b[j])
{
if (b[j]>j-flag) biu();
if (b[j]<j-flag) Ans=Ans*2%P;
flag=j;
}
}
}
}
// cout<<Ans<<endl;
for (int i=1;i<=n;i++)
{
// cout<<i<<' '<<sum[i]<<endl;
if ((i&1)&&i!=1) Ans=(ll)Ans*F(sum[i],i)%P;
else Ans=(ll)Ans*G(sum[i],i)%P;
}
printf("%d\n",Ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#define MOD 1000000007
long long dp[100005], ans = 1;
int pre[100005], nxt[100005], in[100005], cnt[100005];
bool vis[100005];
int main()
{
// freopen("AGC008-E.in", "r", stdin);
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d", nxt + i);
in[--nxt[i]]++;
}
for (int i = 0; i < n; i++)
{
if (in[i] > 2)
{
puts("0");
return 0;
}
if (in[i] != 2 || vis[i])
continue;
int u = i;
do
{
if (vis[u])
{
puts("0");
return 0;
}
vis[u] = true;
pre[nxt[u]] = u;
u = nxt[u];
}
while (u != i);
}
for (int i = 0; i < n; i++)
{
if (in[i])
continue;
int u = i, link_len = 0, cyc_len = 0;
while (!vis[u])
{
vis[u] = true;
u = nxt[u];
link_len++;
}
do
{
u = pre[u];
cyc_len++;
}
while (in[u] == 1);
if (link_len < cyc_len)
ans = ans * 2 % MOD;
else if (link_len > cyc_len)
{
puts("0");
return 0;
}
}
for (int i = 0; i < n; i++)
{
if (vis[i])
continue;
int u = i, len = 0;
do
{
vis[u] = true;
u = nxt[u];
len++;
}
while (u != i);
cnt[len]++;
}
for (int i = 1; i <= n; i++)
{
int coef = ((i & 1) && i != 1) + 1;
dp[0] = 1;
dp[1] = coef;
for (int j = 2; j <= cnt[i]; j++)
dp[j] = (dp[j - 2] * (j - 1) * i + dp[j - 1] * coef) % MOD;
ans = ans * dp[cnt[i]] % MOD;
}
printf("%lld\n", ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
int q=0;char ch=' ';
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
return q;
}
const int mod=1e9+7,N=100005;
int n,ans;
int a[N],du[N],cir[N],vis[N],footL[N],sum[N],f[N];
int qm(int x) {return x>=mod?x-mod:x;}
void workcir(int x) {
int now=0,fr=0,ed=0,frL=0;
while(cir[x]) {
++now,cir[x]=0;
if(footL[x]) {
if(!fr) ed=fr=now,frL=footL[x];
else {
int kl=(footL[x]<now-ed)+(footL[x]<=now-ed);
ans=1LL*ans*kl%mod,ed=now;
}
}
x=a[x];
}
if(!fr) ++sum[now];
else {
int kl=(frL<now-ed+fr)+(frL<=now-ed+fr);
ans=1LL*ans*kl%mod;
}
}
void work() {
for(RI i=1;i<=n;++i) {
if(du[i]) continue;
int x=i,len=0;while(!cir[x]) x=a[x],++len;
footL[x]=len;
}
ans=1;
for(RI i=1;i<=n;++i) if(cir[i]) workcir(i);
//cout<<ans<<endl;
for(RI i=1;i<=n;++i) {
if(!sum[i]) continue;
f[0]=1;
for(RI j=1;j<=sum[i];++j) {
if(i>1&&(i&1)) f[j]=qm(f[j-1]+f[j-1]);
else f[j]=f[j-1];
if(j>1) f[j]=qm(f[j]+1LL*f[j-2]*(j-1)%mod*i%mod);
}
ans=1LL*ans*f[sum[i]]%mod;
}
}
int main()
{
n=read();
for(RI i=1;i<=n;++i) a[i]=read(),++du[a[i]];
for(RI i=1;i<=n;++i) {
if(vis[i]) continue;
int x=i;while(!vis[x]) vis[x]=i,x=a[x];
if(vis[x]!=i) continue;
while(!cir[x]) cir[x]=1,x=a[x];
}
for(RI i=1;i<=n;++i)
if((cir[i]&&du[i]>2)||(!cir[i]&&du[i]>1)) {puts("0");return 0;}
work();
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#define maxn 100005
#define mod 1000000007
int n,a[maxn],deg[maxn],pre[maxn],cnt[maxn],f[maxn],ans=1;
bool vis[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) { scanf("%d",&a[i]); deg[a[i]]++; }
for(int i=1;i<=n;i++)
{
if(deg[i]>2) { printf("0\n"); return 0; }
if(deg[i]<2||vis[i]) continue;
int p=i;
do
{
if(vis[p]) { printf("0\n"); return 0; }
vis[p]=true; pre[a[p]]=p,p=a[p];
}while(p!=i);
}
for(int i=1;i<=n;i++)
if(!deg[i])
{
int p=i,l1=0,l2=0;
while(!vis[p]) vis[p]=true,p=a[p],l1++;
do l2++,p=pre[p]; while(deg[p]!=2);
if(l1<l2) ans=ans*2%mod;
else if(l1>l2) { printf("0\n"); return 0; }
}
for(int i=1;i<=n;i++)
if(!vis[i])
{
int p=i,l=0;
do l++,p=a[p],vis[p]=true; while(p!=i);
cnt[l]++;
}
for(int i=1;i<=n;i++)
{
int mul=1;
if(i!=1&&(i&1)) mul++;
f[0]=1,f[1]=mul;
for(int j=2;j<=cnt[i];j++) f[j]=(1ll*f[j-2]*(j-1)%mod*i%mod+1ll*f[j-1]*mul%mod)%mod;
ans=1ll*ans*f[cnt[i]]%mod;
}
printf("%d\n",ans);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define N 1111116
#define mod 1000000007
int n,a[N],d[N],h[N],g[N];
long long ans,f[N];
queue<int> q;
int read(){
int x=0,f=1;char ch=getchar();
for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
int main(){
n=read();ans=1;
for (int i=1;i<=n;i++) a[i]=read(),d[a[i]]++;
for (int i=1;i<=n;i++) if (!d[i]) q.push(i);
while (!q.empty()){
int t=q.front();q.pop();
if (h[a[t]]) ans=0;h[a[t]]=h[t]+1;
if (--d[a[t]]==0) q.push(a[t]);
}
for (int i=1;i<=n;i++)
if (d[i]){
int t=i;vector<int> q;
while (a[t]!=i&&!h[t]) t=a[t];
while (d[t]){
q.push_back(h[t]);
d[t]=0;t=a[t];
}
if (!q[0]) g[q.size()]++;
else {
q.push_back(q[0]);
for (int i=1,j=0;i<(int)q.size();i++)
if (q[i]){
if (q[i]<i-j) ans=ans*2%mod;
if (q[i]>i-j) ans=0;
j=i;
}
}
}
for (int i=1;i<=n;i++){
f[0]=1;f[1]=(i>1&&i%2)+1;
for (int j=2;j<=g[i];j++)
f[j]=(f[j-1]*f[1]+f[j-2]*(j-1)*i)%mod;
ans=(ans*f[g[i]])%mod;
}
printf("%lld\n",ans);return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100020
#define mod 1000000007
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*f;
}
int n,a[N],du[N],col[N],len[N],cnt[N];
ll ans=1,dp[N];
bool cir[N];
inline int calc(int l1,int l2){
return (l1<l2)+(l1<=l2);
}
int main(){
// freopen("a.in","r",stdin);
n=read();for(int i=1;i<=n;++i) a[i]=read(),du[a[i]]++;
for(int i=1;i<=n;++i){
if(col[i]) continue;int x=i;
for(;!col[x];x=a[x]) col[x]=i;
if(col[x]!=i) continue;//成环了，标记下环上的点
for(;!cir[x];x=a[x]) cir[x]=1;
}for(int i=1;i<=n;++i) if((!cir[i]&&du[i]>1)||(cir[i]&&du[i]>2)){puts("0");return 0;}
for(int i=1;i<=n;++i){//计算foot的长度
if(du[i]) continue;int x=i,res=0;
while(!cir[x]) ++res,x=a[x];len[x]=res;
}for(int i=1;i<=n;++i){//处理基环内向树
if(!cir[i]) continue;int x=i,fir=0,firlen=0,k=0,last=0;
while(cir[x]){
++k;cir[x]=0;
if(len[x]){
if(!fir){fir=last=k;firlen=len[x];x=a[x];continue;}
ans=ans*calc(len[x],k-last)%mod;if(!ans){puts("0");return 0;}last=k;
}x=a[x];
}if(!fir) cnt[k]++;else ans=ans*calc(firlen,fir+k-last)%mod;
}if(!ans){puts("0");return 0;}
for(int i=1;i<=n;++i){//处理大小为i的所有环
if(!cnt[i]) continue;dp[0]=1;int op=(i&1)+(i>1);//奇环且大于1有两种
for(int j=1;j<=cnt[i];++j){
dp[j]=dp[j-1]*op%mod;
if(j>1) dp[j]=(dp[j]+(ll)(j-1)*i*dp[j-2]%mod)%mod;
}ans=ans*dp[cnt[i]]%mod;
}printf("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10,mod=1e9+7;
int n,ans=1;
int t[N],a[N],vs[N],d[N],dr[N],f[N];
bool cir[N];
inline int ad(int x,int y){x+=y;return x>=mod?x-mod:x;}
void sol(int x)
{
int nw=0,fi=0,sc,pre;
for(;cir[x];x=a[x]){
++nw;cir[x]=false;
if(!dr[x]) continue;
if(!fi) {fi=sc=nw;pre=dr[x];}
else{
if(nw-sc<dr[x]) ans=0;
else if(nw-sc>dr[x]) ans=ad(ans,ans);
sc=nw;
}
}
if(!fi) t[nw]++;
else{
nw=fi-sc+nw;
if(nw<pre) ans=0;
else if(nw>pre) ans=ad(ans,ans);
}
}
int main(){
int i,j,k;
scanf("%d",&n);
for(i=1;i<=n;++i) {scanf("%d",&a[i]);d[a[i]]++;}
for(i=1;i<=n;++i) if(!vs[i]){
vs[i]=i;
for(j=a[i];!vs[j];j=a[j]) vs[j]=i;
if(vs[j]^i) continue;
for(;!cir[j];j=a[j]) cir[j]=true;
}
for(i=1;i<=n;++i)
if((cir[i] && d[i]>2)||((!cir[i])&& d[i]>1))
{puts("0");return 0;}
for(i=1;i<=n;++i) if(!d[i]){
for(k=0,j=i;(!cir[j]);j=a[j]) k++;
dr[j]=k;
}
for(i=1;i<=n;++i) if(cir[i]) sol(i);
if(!ans) {puts("0");return 0;}
f[0]=1;
for(i=1;i<=n;++i) if(t[i]){
for(j=1;j<=t[i];++j){
if(i>1 && (i&1)) f[j]=ad(f[j-1],f[j-1]);
else f[j]=f[j-1];
if(j>1) f[j]=ad(f[j],(ll)f[j-2]*(j-1)*i%mod);
}
ans=(ll)ans*f[t[i]]%mod;
}
printf("%d",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define Set(a,b) memset(a,b,sizeof(a))
typedef long long ll;
const int N=1e5+10;
const int mod=1e9+7;
int fa[N],n,son[N],len[N];
bool cir[N];int vis[N];
int num[N];int dp[N];
template<class T>inline void Inc(T&x,int y){x+=y;if(x>=mod) x-=mod;return;}
inline int Mod(int x){if(x>=mod) x-=mod;return x;}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;++i) scanf("%d",&fa[i]),++son[fa[i]];
for(int i=1;i<=n;++i) {
if(vis[i]) continue;
int u=i;
while(!vis[u]) vis[u]=i,u=fa[u];
if(vis[u]==i) {while(!cir[u]) cir[u]=1,u=fa[u];}
}
for(int i=1;i<=n;++i) if((cir[i]&&son[i]>2)||(!cir[i]&&son[i]>1)) return puts("0"),0;
for(int i=1;i<=n;++i) {
if(!son[i]) {
int u=i,l=0;
while(!cir[u]) ++l,u=fa[u];
if(len[u]) return puts("0"),0;
len[u]=l;
}
}int ans=1;
for(int i=1;i<=n;++i){
if(!cir[i]) continue;
int size=0,fir=0,pfir=0,now=0;
for(int u=i;cir[u];u=fa[u]){
++size;cir[u]=0;
if(len[u]) {
if(!fir) fir=size,pfir=u,now=size;
else {
int pre=now;now=size;int d=now-pre;
if(d>len[u]) Inc(ans,ans);
else if(d<len[u]) {puts("0");return 0;}
}
}
}
if(!fir) ++num[size];
else {
int d=size-now+fir;
if(d>len[pfir]) Inc(ans,ans);
else if(d<len[pfir]) {puts("0");return 0;}
}
}
for(int i=1;i<=n;++i){
if(!num[i]) continue;dp[0]=1;
for(int j=1;j<=num[i];++j) {
dp[j]=dp[j-1];//keep the same
if(i>1&&(i&1)) Inc(dp[j],dp[j]);//two kinds
if(j>1) Inc(dp[j],(ll)dp[j-2]*(j-1)%mod*i%mod);// union
}
ans=(ll)ans*dp[num[i]]%mod;
}
printf("%d\n",ans%mod);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
template<typename T>inline T read(){
T f=0,x=0;char c=getchar();
while(!isdigit(c)) f=c=='-',c=getchar();
while(isdigit(c)) x=x*10+c-48,c=getchar();
return f?-x:x;
}
namespace run{
const int N=1e5+9,mod=1e9+7;
inline int add(int x,int y){return x+y>=mod?x-mod+y:x+y;}
int n,a[N],cir[N],vis[N],deg[N],len[N],top[N],sum[N],f[N];
int ans=1;
inline void solve(int x){
int tmp=a[x],las=0,now=0,ret=1;
while(cir[tmp]==1){
now++;
if(len[tmp]){
int res=(now-las>=len[tmp])+(now-las>len[tmp]);
ret=1LL*ret*res%mod,las=now;
}
cir[tmp]=2,tmp=a[tmp];
}
ans=1LL*ans*ret%mod;
}
int main(){
n=read<int>();
for(int i=1;i<=n;i++) deg[a[i]=read<int>()]++;
for(int i=1;i<=n;i++){
int tmp=i;
while(!vis[tmp]) vis[tmp]=i,tmp=a[tmp];
if(vis[tmp]==i){
int now=tmp;
do{cir[now]=1,now=a[now];}while(now!=tmp);
}
}
for(int i=1;i<=n;i++)
if((cir[i] && deg[i]>2) || (!cir[i] && deg[i]>1)) puts("0"),exit(0);
for(int i=1;i<=n;i++) if(!deg[i]){
int tmp=i,siz=0;
while(!cir[tmp]) siz++,tmp=a[tmp];
len[tmp]=siz,top[i]=tmp;
}
for(int i=1;i<=n;i++) if(!deg[i] && cir[top[i]]==1) solve(top[i]);
for(int i=1;i<=n;i++) if(cir[i]==1){
int tmp=i,siz=0;
while(cir[tmp]==1) cir[tmp]=2,siz++,tmp=a[tmp];
sum[siz]++;
}
for(int i=1;i<=n;i++) if(sum[i]){
f[0]=1;
for(int j=1;j<=sum[i];j++){
if(i>1 && (i&1)) f[j]=add(f[j-1],f[j-1]);
else f[j]=f[j-1];
if(j>1) f[j]=(1LL*f[j-2]*i%mod*(j-1)+f[j])%mod;
}
ans=1LL*ans*f[sum[i]]%mod;
}
printf("%d\n",ans);
return 0;
}
}
int main(){
#ifdef my
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif
return run::main();
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | # include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn(1e5 + 5);
const int mod(1e9 + 7);
inline void Inc(int &x, int y) {
x = x + y >= mod ? x + y - mod : x + y;
}
inline void Dec(int &x, int y) {
x = x - y < 0 ? x - y + mod : x - y;
}
inline int Add(int x, int y) {
return x + y >= mod ? x + y - mod : x + y;
}
inline int Sub(int x, int y) {
return x - y < 0 ? x - y + mod : x - y;
}
int n, cir[maxn], cnt, fa[maxn], vis[maxn], len, d[maxn], in[maxn];
int ans, f[maxn], a[maxn], chain[maxn], que[maxn << 1];
inline void GetCircle() {
int i;
for (i = 1; i <= len; ++i) if (d[que[i]] ^ 1) return;
cir[++cnt] = len;
for (i = 1; i <= len; ++i) vis[que[i]] = 2;
}
void Dfs1(int u) {
int cur;
vis[u] = 1, in[u] = 1;
if (!vis[a[u]]) fa[a[u]] = u, Dfs1(a[u]);
else if (in[a[u]]) {
len = 0;
for (cur = u; ; cur = fa[cur]) {
que[++len] = cur, vis[cur] = 3;
if (cur == a[u]) break;
}
GetCircle();
}
in[u] = 0;
}
void Dfs2(int u) {
chain[a[u]] = chain[u] + 1;
if (vis[a[u]] > 1) return;
Dfs2(a[u]);
}
int Solve(int x) {
int cur, i, j, ret = 1;
que[len = 1] = x;
for (cur = a[x]; cur ^ x; cur = a[cur]) que[++len] = cur;
reverse(que + 1, que + len + 1), cur = len + len;
for (i = 1; i <= len; ++i) vis[que[i]] = 4, que[len + i] = que[i];
for (i = 1; i <= len; ++i)
if (chain[que[i]]) {
for (j = i + 1; j <= cur && !chain[que[j]]; ++j);
if (chain[que[i]] > j - i) puts("0"), exit(0);
if (chain[que[i]] < j - i) Inc(ret, ret);
i = j - 1;
}
return ret;
}
int main() {
int i, j, k, ret = 1;
scanf("%d", &n);
for (i = 1; i <= n; ++i) scanf("%d", &a[i]), ++d[a[i]];
for (i = 1; i <= n; ++i) if (!d[i]) Dfs1(i);
for (i = 1; i <= n; ++i) if (!vis[i]) Dfs1(i);
for (i = 1; i <= n; ++i)
if ((vis[i] > 1 && d[i] > 2) || (vis[i] == 1 && d[i] > 1)) return puts("0"), 0;
sort(cir + 1, cir + cnt + 1);
for (i = 1; i <= n; i = j) {
for (j = i; j <= n && cir[j] == cir[i]; ++j);
f[i - 1] = 1;
for (k = i; k < j; ++k) {
f[k] = f[k - 1];
if (cir[i] > 1 && (cir[i] & 1)) Inc(f[k], f[k - 1]);
if (k > i) Inc(f[k], (ll)f[k - 2] * (k - i) % mod * cir[i] % mod);
}
ret = (ll)ret * f[j - 1] % mod;
}
for (i = 1; i <= n; ++i) if (vis[i] == 1) vis[i] = 0;
for (i = 1; i <= n; ++i) if (!d[i]) Dfs2(i);
for (i = 1; i <= n; ++i) if (vis[i] == 3) ret = (ll)ret * Solve(i) % mod;
printf("%d\n", ret);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <iostream>
#include <cstdio>
#include <cstring>
#define N 100005
using namespace std;
typedef long long ll;
const int mod=1e9+7;
int n,a[N],D,IN[N],siz,sum[N],len[N],b[N];
bool inc[N];
ll f[N];
bool vis[N],used[N];
ll calc(ll x){
return x*(x-1)/2%mod;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),IN[a[i]]++;
for(int i=1;i<=n;i++){
if(vis[i]) continue;
int g=i;
while(!vis[g]) vis[g]=1,g=a[g];
if(used[g]){
g=i;
while(!used[g]) used[g]=1,g=a[g];
continue;
}
int x=a[g];inc[g]=1;b[a[g]]=g;
while(x!=g) inc[x]=1,b[a[x]]=x,x=a[x];
g=i;
while(!used[g]) used[g]=1,g=a[g];
}
for(int i=1;i<=n;i++){
if(inc[i]){if(IN[i]>2) return puts("0"),0;}
else if(IN[i]>1) return puts("0"),0;
}
for(int i=1;i<=n;i++){
if(!IN[i]){
int g=i,l=0;
while(!inc[g]) l++,g=a[g];
len[g]=l;
}
}
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++){
if(!vis[i]&&inc[i]){
int g=i,t=1,L=0;
if(len[i]) t=0;
while(!vis[g]){
if(len[g]) t=0;
L++;
vis[g]=1;
g=a[g];
}
if(t) sum[L]++;
}
}
ll ans=1;
for(int i=1;i<=n;i++){
ll g=1+(i&1);
if(i==1) g--;
f[0]=1;
for(int p=1;p<=sum[i];p++){
f[p]=f[p-1]*g;
if(p>1) f[p]=(f[p]+f[p-2]*(p-1)%mod*i)%mod;
}
ans=ans*f[sum[i]]%mod;
}
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++){
if(inc[i]&&len[i]){
int g=i;
int x=b[g],L=1;
while(!len[x]) x=b[x],L++;
if(len[g]>L) return puts("0"),0;
if(len[g]<L) ans=ans*2%mod;
g=a[g];L=1;
while(g!=i){
if(!len[g]){
L++;g=a[g];continue;
}
if(len[g]>L) return puts("0"),0;
if(len[g]<L)
ans=ans*2%mod;
L=1;g=a[g];
}
g=i;
while(inc[g]) inc[g]=0,g=a[g];
}
}
printf("%lld",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
using namespace std;
typedef long long LL;
const int maxn = 200005;
const int mod = 1e9+7;
bool vis[maxn],ins[maxn];int n,a[maxn],deg[maxn];int ans=1;
bool onring[maxn];int stk[maxn],top,len[maxn],cnt[maxn];
int fac[maxn],inc[maxn],pw[maxn],ffac[maxn];
int fpm(LL p,int k) {
LL res=1ll;
while (k) {
if (k&1) (res*=p)%=mod;
(p*=p)%=mod;k>>=1;
}
return res;
}
LL comb(int n,int m) {
return (LL)fac[n]*inc[m]%mod*(LL)inc[n-m]%mod;
}
void dfs(int u)
{
stk[++top]=u;ins[u]=vis[u]=true;
if (ins[a[u]]) {
onring[a[u]]=true;
for (int j=top;stk[j]!=a[u];j--)
onring[stk[j]]=true;
}
if (!vis[a[u]]) dfs(a[u]);
top--;ins[u]=false;
}
void dfs2(int u,int dis)
{
if (onring[u]) len[u]=dis;
else dfs2(a[u],dis+1);
}
void calc(int u)
{
vis[u]=true;
for (int v=a[u];v!=u;v=a[v]) vis[v]=true;
if (!len[u]) {
int v=a[u],dis=1;
while (v!=u&&!len[v]) {v=a[v],++dis;}
if (v==u) {cnt[dis]++;return ;}
u=v;
}
for (int co=0,i=u,j;!co||i!=u;co++,i=j)
{
int dis=1;j=a[i];while (!len[j]) j=a[j],++dis;
if (dis<len[j]) {puts("0");exit(0);}
if (dis>len[j]) (ans<<=1)%=mod;
}
}
int main()
{
#ifdef Amberframe
freopen("agc008e.in","r",stdin);
freopen("agc008e.out","w",stdout);
#endif
scanf("%d",&n);fac[0]=inc[0]=ffac[0]=pw[0]=1;
for (int i=1;i<=n;i++) scanf("%d",&a[i]),deg[a[i]]++;
for (int i=1;i<=n;i++) if (!vis[i]) dfs(i);
for (int i=1;i<=n;i++) if (onring[i]+1<deg[i]) {puts("0");return 0;}
for (int i=1;i<=n;i++) if (!deg[i]) dfs2(i,0);
memset(vis,0,sizeof vis);
for (int i=1;i<=n;i++) if (onring[i]&&!vis[i]) calc(i);
for (int i=1;i<=n;i++) ffac[i]=(LL)(2*i-1)*ffac[i-1]%mod;
for (int i=1;i<=n;i++) fac[i]=(LL)fac[i-1]*i%mod;
inc[n]=fpm(fac[n],mod-2);
for (int i=n;i>=2;i--) inc[i-1]=(LL)inc[i]*i%mod;
for (int i=1;i<=n;i++) pw[i]=(LL)pw[i-1]*2ll%mod;
for (int i=1;i<=n;i++)
if (i&1) {
int o=cnt[i],res=0,lx=1;
for (int x=0;x*2<=o;x++,lx=(LL)lx*i%mod)
res=(res+(LL)comb(o,x*2)*ffac[x]%mod*(LL)lx%mod*(i>1?pw[o-2*x]:1))%mod;
ans=(LL)ans*res%mod;
}
else {
int o=cnt[i],res=0,lx=1;
for (int x=0;x*2<=o;x++,lx=(LL)lx*i%mod)
res=(res+(LL)comb(o,x*2)*ffac[x]%mod*(LL)lx)%mod;
ans=(LL)ans*res%mod;
}
printf("%d",ans<0?ans+mod:ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod=1000000007,N=100005;
bool in[N];
int vis[N],a[N],du[N],len[N],num[N],f[N];
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// ios::sync_with_stdio(false);
int n,ans=1;
cin>>n;
for(int i=1;i<=n;i++)
{
// cin>>a[i];
scanf("%lld",&a[i]);
du[a[i]]++;
}
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
int p=i;
while(!vis[p])
{
vis[p]=i;
p=a[p];
}
if(vis[p]==i)
{
while(!in[p])
{
in[p]=true;
p=a[p];
}
}
}
}
// return 0;
for(int i=1;i<=n;i++)
{
// cout<<"W "<<i<<" "<<p<<endl;
if(du[i]==0)
{
int p=i,res=0;
while(!in[p])
{
++res;
p=a[p];
// if(res>100000) break;
}
//cout<<"N "<<p<<" "<<in[11]<<endl;
len[p]=res;
}
}
// return 0;
for(int i=1;i<=n;i++)
{
if((in[i]&&du[i]>2)||(in[i]==false&&du[i]>1))
{
cout<<0;
return 0;
}
}
// return 0;
for(int i=1;i<=n;i++)
{
if(in[i])
{
int p=i,res=0,pos=0,sum;
while(in[p])
{
++res;
if(len[p])
{
if(!pos)
{
pos=p;
sum=res;
}
else
{
// cout<<"P "<<p<<" "<<len[p]<<" "<<res<<endl;
if(res>len[p]) ans=ans*2%mod;
else if(res<len[p]) ans=0;
}
res=0;
}
in[p]=false;
p=a[p];
}
if(pos)
{
res+=sum;
// cout<<"Q "<<pos<<" "<<len[pos]<<" "<<res<<endl;
if(res>len[pos]) ans=ans*2%mod;
else if(res<len[pos]) ans=0;
}
else num[res]++;
}
}
// cout<<ans<<endl;
for(int i=1;i<=n;i++)
{
if(num[i])
{
f[0]=1;
for(int j=1;j<=num[i];j++)
{
if(i%2==1&&i!=1) f[j]=f[j-1]*2%mod;
else f[j]=f[j-1];
if(j-2>=0) f[j]=(f[j]+(j-1)*i%mod*f[j-2]%mod)%mod;
}
// cout<<i<<" "<<f[num[i]]<<endl;
ans=ans*f[num[i]]%mod;
}
}
cout<<ans;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
int ad(int x,int y) { x+=y; return x>=mod?x-mod:x; }
int sb(int x,int y) { x-=y; return x<0?x+mod:x; }
int mu(int x,int y) { return 1ll*x*y%mod; }
void up(int &x,int y) { x+=y; if(x>=mod)x-=mod; }
int ksm(int a,int b) {
int ans = 1;
for(;b;b>>=1,a=mu(a,a))
if(b&1) ans=mu(ans,a);
return ans;
}
int n;
const int maxn = 2e5+5;
int a[maxn],rd[maxn],from[maxn],f[maxn];
bool vis[maxn];
int ans = 1;
int CR[maxn],cnt;
int main() {
scanf("%d",&n);
for(int i=1;i<=n;i++) {
scanf("%d",&a[i]);
rd[a[i]]++;
}
for(int i=1;i<=n;i++) {
if(rd[i]>2) { puts("0"); return 0; }
else if(rd[i]<2||vis[i]) continue;
int p = i;
do{
if(vis[p]) { puts("0"); return 0; }
vis[p] = 1; from[a[p]] = p;
p = a[p];
}while(p!=i);
}
for(int i=1;i<=n;i++) {
if(!rd[i]) {
int p = i; int l1 = 0; int l2 = 0;
while(!vis[p]) vis[p] = 1 , l1++ , p = a[p];
do{
l2++; p = from[p];
}while(rd[p]!=2);
if(l1==l2) continue;
else if(l1<l2) ans = mu(ans,2);
else { puts("0"); return 0; }
}
}
for(int i=1;i<=n;i++) {
if(!vis[i]) {
int p = i; int cd = 0;
do{
p = a[p];
cd++;
vis[p] = 1;
}while(p!=i);
CR[++cnt] = cd;
}
}
sort(CR+1,CR+1+cnt);
for(int i=1,j;i<=n;i=j+1) {
j = i;
while(j<n&&CR[j+1]==CR[i]) j++;
f[i-1] = 1;
for(int k=i;k<=j;k++) {
f[k] = f[k-1];
if(CR[i]>1&&(CR[i]&1)) up(f[k],f[k-1]);
if(k!=i) up(f[k],mu(f[k-2],mu(k-i,CR[i]) ) );
}
ans = mu(ans,f[j]);
}
printf("%d",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long s64;
inline int getint()
{
static char c;
while ((c = getchar()) < '0' || c > '9');
int res = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
res = res * 10 + c - '0';
return res;
}
const int MaxN = 100000;
const int M = 1000000007;
inline int modpow(int a, const int &n)
{
int res = 1;
for (int i = n; i; i >>= 1)
{
if (i & 1)
res = (s64)res * a % M;
a = (s64)a * a % M;
}
return res;
}
int n;
int link[MaxN + 1];
int fact[MaxN + 1];
int rfact[MaxN + 1];
int prePow[MaxN + 1];
inline int binom(const int &n, const int &m)
{
return (s64)fact[n] * rfact[m] % M * rfact[n - m] % M;
}
int dep[MaxN + 1];
int deg[MaxN + 1];
int ori[MaxN + 1];
int q_n, q[MaxN];
bool vis[MaxN + 1];
int cir_l, cir[MaxN + 1];
int sum[MaxN + 1];
int main()
{
cin >> n;
prePow[0] = 1;
for (int i = 1; i <= n; ++i)
prePow[i] = prePow[i - 1] * 2 % M;
fact[0] = 1;
for (int i = 1; i <= n; ++i)
fact[i] = (s64)fact[i - 1] * i % M;
rfact[n] = modpow(fact[n], M - 2);
for (int i = n; i; --i)
rfact[i - 1] = (s64)rfact[i] * i % M;
for (int u = 1; u <= n; ++u)
{
int v = getint();
++deg[link[u] = v];
++ori[v];
}
q_n = 0;
for (int u = 1; u <= n; ++u)
if (!deg[u])
q[q_n++] = u;
for (int i = 0; i < q_n; ++i)
{
int u = q[i];
dep[link[u]] = dep[u] + 1;
if (!--deg[link[u]])
q[q_n++] = link[u];
}
for (int u = 1; u <= n; ++u)
if (ori[u] - deg[u] > 1)
{
cout << 0 << endl;
return 0;
}
int res = 1;
for (int u = 1; u <= n; ++u)
if (!vis[u] && deg[u])
{
cir_l = 0;
int sv = u;
bool is_single = true;
while (!vis[sv])
{
if (dep[sv] > 0)
is_single = false;
vis[sv] = true;
cir[++cir_l] = sv;
sv = link[sv];
}
if (is_single)
{
++sum[cir_l];
continue;
}
for (int i = 1; i <= cir_l; ++i)
{
int sv = cir[i];
if (dep[sv] > 0)
{
int pos = i, len = 0;
for (int k = 1; ; ++k)
{
pos = pos == 1 ? cir_l : pos - 1;
if (dep[cir[pos]] > 0)
{
len = k;
break;
}
}
if (len > dep[sv])
res = res * 2 % M;
else if (len < dep[sv])
{
cout << 0 << endl;
return 0;
}
}
}
}
for (int l = 1; l <= n; ++l)
if (sum[l] > 0)
{
s64 p = 1, r = 0;
for (int i = 0; i <= sum[l] / 2; ++i)
{
s64 ways = (s64)p * fact[i] % M;
ways = (s64)ways * binom(sum[l], i * 2) % M;
ways = (s64)ways * binom(i * 2, i) % M;
ways = (s64)ways * modpow(prePow[i], M - 2) % M;
if (l > 1 && (l & 1))
ways = (s64)ways * prePow[sum[l] - i * 2] % M;
r = (r + ways) % M;
p = (s64)p * l % M;
}
res = (s64)res * r % M;
}
cout << res << endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
vector<int>A[110000];
int n,x[110000],pd[110000],len,B[110000],num[110000],f[110000];
map<long long,int>M;
const int mo=1e9+7;
void dfs(int k){
if (pd[k]) return;
pd[k]=1; B[++len]=k;
dfs(x[k]);
for (int i=0;i<A[k].size();i++) dfs(A[k][i]);
}
int getans(int k1,int k2,int k3){
long long st=1ll*k1*n+k2;
if (M.count(st)) return M[st];
int ans=0;
if (A[k1].size()+A[k2].size()>2||A[k1].size()>1||(A[k1].size()==0&&A[k2].size()==0)){
M[st]=0; return 0;
}
int flag=0;
for (int i=0;i<A[k1].size();i++) if (A[k1][i]==k3) flag=1;
for (int i=0;i<A[k2].size();i++) if (A[k2][i]==k3) flag=2;
if (flag==1){
ans=1;
if (A[k2].size()>0&&A[A[k2][0]].size()>0) ans=0;
// cout<<"getans1 "<<k1<<" "<<k2<<" "<<k3<<" "<<ans<<endl;
M[st]=ans; return ans;
}
if (flag==2){
int ex=0; ans=1;
if (A[k1].size()>0) ex=A[k1][0];
if (ex) ans=0;
if (A[k2].size()>1) ex=A[k2][0]+A[k2][1]-k3;
if (ex&&A[ex].size()>0) ans=0;
// cout<<"getans2 "<<k1<<" "<<k2<<" "<<k3<<" "<<ans<<endl;
M[st]=ans; return ans;
}
if (A[k1].size()==0){
if (A[k2].size()==1){
ans=(ans+getans(0,A[k2][0],k3))%mo;
} else {
ans=(ans+getans(A[k2][0],A[k2][1],k3))%mo;
ans=(ans+getans(A[k2][1],A[k2][0],k3))%mo;
}
} else if (A[k2].size()==0)
ans=(ans+getans(0,A[k1][0],k3))%mo;
else ans=(ans+getans(A[k1][0],A[k2][0],k3))%mo;
M[st]=ans;
// cout<<"getans "<<k1<<" "<<k2<<" "<<k3<<" "<<ans<<endl;
return ans;
}
int solve(){
int flag=0,where=0; M.clear();
for (int i=1;i<=len;i++) if (A[B[i]].size()!=1) flag=1;
//for (int i=1;i<=len;i++) cout<<B[i]<<" "; cout<<endl;
if (flag==0){
num[len]++; return 1;
}
for (int i=1;i<=len;i++) if (A[B[i]].size()==2) where=B[i];
int ans=getans(0,where,where);
return ans;
}
int main(){
scanf("%d",&n);
for (int i=1;i<=n;i++) scanf("%d",&x[i]),A[x[i]].push_back(i);
for (int i=1;i<=n;i++) if (A[i].size()>2){
printf("0\n"); return 0;
}
int ans=1;
for (int i=1;i<=n;i++) if (pd[i]==0){
len=0; dfs(i);
ans=1ll*ans*solve()%mo;
}
//cout<<ans<<endl;
for (int i=1;i<=n;i++){
f[0]=1; f[1]=1;
if ((i&1)&&(i>1)) f[1]=2;
for (int j=2;j<=num[i];j++)
f[j]=(1ll*f[j-2]*(j-1)%mo*i+1ll*f[j-1]*(1+(i&1)-(i==1)))%mo;
ans=1ll*ans*f[num[i]]%mo;
}
cout<<ans<<endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i)
const int N = 100005, M = N << 1, mo = 1e9 + 7 ;
using namespace std ;
int n, e, ter[M], nxt[M], lnk[N], opt[N], h[N], ans, d[N], q[N], f[N], deg[N] ;
bool vis[N] ;
void add(int x, int y) {
ter[++ e] = y, nxt[e] = lnk[x], lnk[x] = e ;
}
bool check(int x) {
vis[x] = false ;
int he = 0, ta = 1 ;
opt[1] = x ;
for ( ; he != ta ; ) {
++ he ;
int u = opt[he] ;
for (int i = lnk[u] ; i; i = nxt[i]) {
int v = ter[i] ;
if (vis[v]) {
vis[v] = false ;
opt[++ ta] = v ;
}
}
}
rep(i, 1, ta) {
int u = opt[i] ;
for (int j = lnk[u]; j; j = nxt[j]) if (j & 1) {
int v = ter[j] ;
++ deg[v] ;
}
}
bool flg = true ;
rep(i, 1, ta) {
int u = opt[i] ;
flg &= deg[u] > 0 ;
}
if (flg) {
++ h[ta] ;
return true ;
}
int heq = 0, taq = 0 ;
rep(i, 1, ta) {
int u = opt[i] ;
d[u] = deg[u] ;
if (!d[u]) q[++ taq] = u ;
if (d[u] > 2) return false ;
}
for ( ; heq != taq ; ) {
++ heq ;
int u = q[heq] ;
for (int i = lnk[u]; i; i = nxt[i]) if (i & 1) {
int v = ter[i] ;
-- d[v] ;
if (!d[v]) q[++ taq] = v ;
}
}
rep(i, 1, ta) {
int u = opt[i] ;
if (!d[u] && deg[u] > 1) return false ;
}
heq = taq = 0 ;
rep(i, 1, ta) {
int u = opt[i] ;
if (!deg[u]) q[++ taq] = u, f[u] = 0 ;
}
for ( ; heq != taq ; ) {
++ heq ;
int u = q[heq] ;
for (int i = lnk[u]; i; i = nxt[i]) if (i & 1) {
int v = ter[i] ;
-- deg[v] ;
f[v] = f[u] + 1 ;
if (!deg[v]) q[++ taq] = v ;
}
}
rep(i, 1, ta) {
int u = opt[i] ;
if (deg[u]) {
x = u ;
break ;
}
}
d[0] = 0 ;
d[++ d[0]] = x ;
for ( ; ; ) {
int u = d[d[0]] ;
for (int i = lnk[u]; i; i = nxt[i]) if (i & 1) {
int v = ter[i] ;
d[++ d[0]] = v ;
break ;
}
if (d[d[0]] == d[1]) {
-- d[0] ;
break ;
}
}
int las = 0 ;
rep(i, 1, d[0]) if (f[d[i]]) las = i ;
rep(i, 1, d[0]) if (f[d[i]]) {
int l1 = f[d[i]], l2 = las >= i ? d[0] - las + i : i - las ;
if (l1 < l2) ans = ans * 2 % mo ; else
if (l1 > l2) ans = 0 ;
las = i ;
}
return true ;
}
inline void upd(int &x, int y) {
x = (x + y) % mo ;
}
int main() {
scanf("%d", &n) ;
int x ;
rep(i, 1, n) {
scanf("%d", &x) ;
add(i, x), add(x, i) ;
}
rep(i, 1, n) vis[i] = true ;
ans = 1 ;
rep(i, 1, n) if (vis[i]) {
if (!check(i)) {
printf("0\n") ;
return 0 ;
}
}
rep(L, 1, n) if (h[L]) {
f[0] = 1 ;
rep(i, 1, h[L]) {
f[i] = 0 ;
upd(f[i], f[i - 1]) ;
if ((L > 1) && (L & 1)) upd(f[i], f[i - 1]) ;
if (i > 1) upd(f[i], (ll) (i - 1) * f[i - 2] % mo * L % mo) ;
}
ans = (ll) ans * f[h[L]] % mo ;
}
printf("%d\n", ans) ;
return 0 ;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <algorithm>
#include <vector>
const int MOD = 1000000007;
int N, a[100001], v[100001], bel[100001], dep[100001], pure[100001], O = 1;
std::vector < int > V[100001];
std::vector < std::vector < int > > cirs;
int main()
{
scanf("%d", &N);
for (int i = 1; i <= N; i++)
scanf("%d", a + i);
for (int i = 1; i <= N; i++)
if (!v[i])
{
static int way[100001];
int L = 1;
v[way[L] = i] = 1;
while (!v[a[way[L]]])
{
v[way[L + 1] = a[way[L]]] = 1;
L++;
}
way[L + 1] = a[way[L]];
if (v[a[way[L]]] == 2)
for (int i = L; i; i--)
{
bel[way[i]] = bel[way[i + 1]];
dep[way[i]] = dep[way[i + 1]] + 1;
}
else
{
int S = std::find(way + 1, way + L + 1, way[L + 1]) - way;
for (int i = S; i <= L; i++)
{
bel[way[i]] = way[i];
dep[way[i]] = 0;
}
for (int i = S - 1; i; i--)
{
bel[way[i]] = bel[way[i + 1]];
dep[way[i]] = dep[way[i + 1]] + 1;
}
cirs.push_back({ way + S, way + L + 1 });
}
for (int i = 1; i <= L; i++)
v[way[i]] = 2;
}
for (int i = 1; i <= N; i++)
V[bel[i]].push_back(dep[i]);
for (int i = 1; i <= N; i++)
{
std::sort(V[i].begin(), V[i].end());
int S = V[i].size();
for (int j = 0; j < S; j++)
if (V[i][j] != j)
{
puts("0");
return 0;
}
}
for (auto &cir : cirs)
{
for (int &i : cir)
i = V[i].size() - 1;
if (int(std::count(cir.begin(), cir.end(), 0)) == int(cir.size()))
pure[cir.size()]++;
else
{
int *A = cir.data(), L = cir.size();
for (int i = 0; i < L; i++)
if (A[i])
{
for (int j = 1; j < A[i]; j++)
if (A[(i - j % L + L) % L])
{
puts("0");
return 0;
}
if (!A[(i - A[i] % L + L) % L])
O = (O + O) % MOD;
}
}
}
for (int i = 1; i <= N; i++)
{
static int f[100001];
f[0] = 1;
for (int j = 1; j <= pure[i]; j++)
f[j] = (f[j - 1] * (i > 1 && (i & 1) ? 2 : 1) + (long long)f[j - 2] * (j - 1) * i) % MOD;
O = (long long)O * f[pure[i]] % MOD;
}
printf("%d\n", O);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
const int MaxN=100010,mod=1000000007;
int N,a[MaxN];
struct edge{int to;edge*next;}E[MaxN],*fir[MaxN];
int C[MaxN],pos[MaxN],dep[MaxN],cnt[MaxN],ways[MaxN];
bool vis[MaxN];
int solve(){
int res=1;
for(int i=0;i<N;i++)if(!vis[i]){
int len=0,tot=0,j=i,k;
for(;!vis[j];j=a[j])vis[j]=1;
for(k=i;vis[k];k=a[k])vis[k]=0;
for(;!vis[j];j=a[j])vis[C[len++]=j]=1;
for(j=0,k=len-1;j<k;j++,k--){
int t=C[j];C[j]=C[k],C[k]=t;
}
for(j=0;j<len;j++){
int x=C[j],d=0;
for(;;){
int c=0,t;
for(edge*e=fir[x];e;e=e->next)if(!vis[e->to])c++,t=e->to;
if(c>1)return 0;
if(!c)break;
vis[x=t]=1;d++;
}
if(d)pos[tot]=j,dep[tot++]=d;
}
if(tot){
for(j=0;j<tot;j++){
int d=(pos[(j+1)%tot]-pos[j])%len;
if(d<=0)d+=len;
if(d<dep[j])return 0;
if(d>dep[j])res=res*2%mod;
}
}
else cnt[len]++;
}
for(int i=1;i<=N;i++){
ways[0]=1;
for(int j=1;j<=cnt[i];j++){
ways[j]=ways[j-1]*(i>1&&i%2?2:1)%mod;
if(j>1)ways[j]=(ways[j]+(j-1ll)*ways[j-2]%mod*i)%mod;
}
res=1ll*res*ways[cnt[i]]%mod;
}
return res;
}
int main(){
while(scanf("%d",&N)==1){
for(int i=0;i<N;i++)scanf("%d",a+i),--a[i],fir[i]=0,vis[i]=0;
for(int i=0;i<N;i++)E[i]=(edge){i,fir[a[i]]},fir[a[i]]=E+i;
for(int i=N;i;i--)cnt[i]=0;
printf("%d\n",solve());
}
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int read(){
int x=0;
char ch=getchar();
while (!isdigit(ch))
ch=getchar();
while (isdigit(ch))
x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
return x;
}
const int N=100005,mod=1e9+7;
int n,a[N],in[N],size[N],vis[N],cro[N];
int q[N],head,tail;
int k[N],m;
int r[N];
void add(int &x,int y){
if ((x+=y)>=mod)
x-=mod;
}
int sc=0;
int Pre(int k,int n){
return k==1?n:(k-1);
}
int solve(int n){
int ans=1,f=0;
// printf("solve :: n = %d ",n);
// for (int i=1;i<=n;i++)
// printf(" %d",k[i]);
// puts("");
for (int i=1;i<=n;i++){
if (k[i]==0)
continue;
f=1;
int x=Pre(i,n);
while (!k[x])
x=Pre(x,n);
int t=(i-x+n)%n;
if (t==0)
t+=n;
// printf("%d :: t = %d\n",i,t);
if (t<k[i])
return 0;
if (t>k[i])
ans=2LL*ans%mod;
}
if (!f)
r[n]++;
return ans;
}
int dp[N];
int main(){
n=read();
memset(in,0,sizeof in);
for (int i=1;i<=n;i++){
in[a[i]=read()]++;
size[i]=1;
}
head=tail=0;
for (int i=1;i<=n;i++){
if (in[i]>2)
return puts("0"),0;
if (in[i]==0)
q[++tail]=i;
if (in[i]==2)
cro[i]=1;
}
while (head!=tail){
int x=q[++head],y=a[x];
size[y]+=size[x];
if (!(--in[y]))
q[++tail]=y;
}
for (int i=1;i<=n;i++)
if (cro[i]&&!in[i])
return puts("0"),0;
int ans=1;
memset(vis,0,sizeof vis);
for (int i=1;i<=n;i++){
if (in[i]==0||vis[i])
continue;
m=0;
for (int x=i;!vis[x];x=a[x])
vis[x]=1,k[++m]=size[x]-1;
ans=1LL*ans*solve(m)%mod;
}
for (int i=1;i<=n;i++)
if (r[i]>0){
dp[0]=1,dp[1]=i>1&&(i&1)?2:1;
for (int j=2;j<=r[i];j++)
dp[j]=(1LL*(j-1)*i%mod*dp[j-2]+dp[j-1]*(i>1&&(i&1)?2:1))%mod;
ans=1LL*ans*dp[r[i]]%mod;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int MaxN=1e5+5;
const int Mod=1e9+7;
int N,Ans=1;
int A[MaxN],Deg[MaxN];
int Pre[MaxN],Cnt[MaxN],F[MaxN];
bool Vis[MaxN];
int main(){
int i,j,l,l1,l2;
scanf("%d",&N);
for(i=1;i<=N;i++)
scanf("%d",&A[i]),Deg[A[i]]++;
for(i=1;i<=N;i++){
if(Deg[i]>2){
puts("0");
return 0;
}
if(Deg[i]<2||Vis[i])
continue;
j=i;
do{
if(Vis[j]){
puts("0");
return 0;
}
Vis[j]=true;
Pre[A[j]]=j;
j=A[j];
}while(i^j);
}
for(i=1;i<=N;i++)
if(!Deg[i]){
l1=l2=0;
for(j=i;!Vis[j];j=A[j])
Vis[j]=true,l1++;
do j=Pre[j],l2++;
while(Deg[j]==1);
if(l1>l2){
puts("0");
return 0;
}
if(l1<l2)
Ans=2*Ans%Mod;
}
for(i=1;i<=N;i++)
if(!Vis[i]){
l=0;
for(j=i;!Vis[j];j=A[j])
Vis[j]=true,l++;
Cnt[l]++;
}
for(i=1;i<=N;i++){
l=((i&1)&&i>1)+1;
F[0]=1,F[1]=l;
for(j=2;j<=Cnt[i];j++)
F[j]=(1ll*F[j-2]*(j-1)*i+F[j-1]*l)%Mod;
Ans=1ll*Ans*F[Cnt[i]]%Mod;
}
printf("%d\n",Ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mxn=1000010,md=1000000007;
int power(int x,int p,int num=1){
for (;p;p>>=1,x=1ll*x*x%md)
if (p&1) num=1ll*num*x%md;
return num;
}
int ans,c[mxn],stk[mxn],tp,son[mxn],vis[mxn],cnt[mxn];
int solve_tr(){
int lst=0;
for (int i=1;i<=tp;++i) cnt[i]=0;
for (int i=1;i<=tp;++i)
for (int x=son[stk[i]];x;++cnt[i],vis[x]=1,x=son[x]);
for (int i=1;i<=tp;++i)
if (cnt[i]) {lst=i;break;}
if (!lst) return ++c[tp],1;
int ret=1;
for (int i=tp;i;++lst,--i)
if (cnt[i]){
if (lst>cnt[i]) ret=ret*2%md;
if (lst<cnt[i]) return 0;
lst=0;
}
return ret;
}
int f[mxn],pw[mxn],fct[mxn],ifc[mxn];
int cal(int x){
return (x-1ll)*x/2%md;
}
int solve_lp(int n,int k){
f[0]=1;
bool flg=((k&1)&&k>1);
int ret=pw[flg*n];
for (int i=1;i<<1<=n;++i){
f[i]=1ll*f[i-1]*cal(n-(i<<1)+2)%md*k%md;
ret=(ret+1ll*f[i]*ifc[i]%md*pw[flg*(n-(i<<1))])%md;
}
return ret;
}
void init(int n){
pw[0]=1;
for (int i=1;i<=n;++i) pw[i]=pw[i-1]*2%md;
fct[0]=ifc[0]=1;
for (int i=1;i<=n;++i) fct[i]=1ll*fct[i-1]*i%md;
ifc[n]=power(fct[n],md-2);
for (int i=n-1;i;--i) ifc[i]=(i+1ll)*ifc[i+1]%md;
}
int n,fa[mxn],stkk[mxn],tpp,tag[mxn],exist[mxn];
bool init2(){
for (int i=1;i<=n;++i)
if (!vis[i]){
tpp=0;
for (int x=i;x;x=fa[x])
if (vis[x]){
if (!exist[x]) break;
for (;stkk[tpp+1]!=x;tag[stkk[tpp]]=1,exist[stkk[tpp--]]=0);
break;
}
else vis[x]=exist[x]=1,stkk[++tpp]=x;
for (;tpp;exist[stkk[tpp--]]=0);
}
memset(vis,0,sizeof(vis));
for (int i=1;i<=n;++i)
if (!tag[i]){
if (son[fa[i]]) return 0;
son[fa[i]]=i;
}
return 1;
}
int main()
{
scanf("%d",&n);
init(n);
for (int i=1;i<=n;++i) scanf("%d",&fa[i]);
if (!init2()) return puts("0"),0;
int ans=1;
for (int i=1;i<=n;++i)
if (tag[i]&&!vis[i]){
tp=tpp=0;
for (int x=i;x;x=fa[x])
if (vis[x]){
for (;stkk[tpp+1]!=x;stk[++tp]=stkk[tpp--]);
break;
}
else vis[x]=1,stkk[++tpp]=x;
ans=1ll*ans*solve_tr()%md;
}
for (int i=1;i<=n;++i)
ans=1ll*ans*solve_lp(c[i],i)%md;
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)n; ++i)
int an = 1, cy[100001], md = 1e9 + 7, pairs[100001], fact[100001], invf[100001], inv[100001];
pair<int, int> y[100001], ln[100001];
bool vis[100001];
void no(){
printf("0\n");
exit(0);
}
int V;
bool go(int v, int m = 0, int d = 0){
if (vis[v]){
if (v == V) { ln[m].first = d; return true; }
return false;
}
vis[v] = true;
if (!y[v].first) { ln[m].second = d; return false; }
bool k = go(y[v].first, y[v].second ? v : m, d + 1);
if (y[v].second)ln[m].first = d, k |= go(y[v].second, v, d + 1);
return k;
}
inline int ch(int n, int r) { return (ll)fact[n] * invf[r] % md * invf[n - r] % md; }
int pw(int x, int p){
if (!p)return 1;
int t = pw(x, p >> 1);
t = (ll)t * t % md;
if (p & 1)t = (ll)t * x % md;
return t;
}
inline void ad(int s, int l){
int a = 0, z = (s & 1 && s != 1) + 1;
for (int i = 0; i <= l; i += 2){
a += (ll)pw(z, l - i) * ch(l, i) % md * pw(s, i >> 1) % md * pairs[i] % md;
if (a >= md)a -= md;
}
an = (ll)an * a % md;
}
int main(){
int n;
scanf("%d", &n);
fact[0] = fact[1] = inv[1] = invf[1] = pairs[0] = invf[0] = 1;
f(i, 2, n + 1){
fact[i] = (ll)fact[i - 1] * i % md;
inv[i] = md - md / i * (ll)inv[md % i] % md;
invf[i] = (ll)invf[i - 1] * inv[i] % md;
if (!(i & 1))pairs[i] = (ll)pairs[i - 2] * (i - 1) % md;
}
f(i, 1, n + 1){
int t;
scanf("%d", &t);
if (y[t].first){
if (y[t].second)no();
y[t].second = i;
}
else y[t].first = i;
}
f(i, 1, n + 1)if (y[i].second){
V = i;
if (!vis[i] && !go(i))no();
if (!ln[i].second)no();
if (ln[i].first > ln[i].second){ if ((an <<= 1) >= md)an -= md; }
else if (ln[i].first < ln[i].second)no();
}
f(i, 1, n + 1)if (!vis[i]){
V = i;
if (go(i, i))++cy[ln[i].first];
}
f(i, 1, n + 1)if (cy[i])ad(i, cy[i]);
printf("%d\n", an);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long i64;
class no_solution: public std::exception
{
public:
no_solution()
{
}
const char * what() const noexcept
{
return "no solution under such constrains";
}
};
const int MAXN = 100000 + 5;
const int MOD = 1e9 + 7;
int N;
i64 answer, dp[MAXN];
int degree[MAXN], lenCnt[MAXN];
int pre[MAXN], nxt[MAXN];
bool vis[MAXN];
void findCycle()
{
for (int i = 1; i <= N; i++)
{
if (degree[i] > 2)
throw no_solution();
if (degree[i] != 2 || vis[i])
continue;
int cur = i;
do
{
if (vis[cur])
throw no_solution();
vis[cur] = true;
pre[nxt[cur]] = cur;
cur = nxt[cur];
} while (cur != i);
}
}
void processCircleBasedTree()
{
for(int i = 1; i <= N; i++)
{
if(degree[i])
continue;
int cur = i, footLen = 0, cycleLen = 0;
while(!vis[cur])
{
vis[cur] = true;
cur = nxt[cur];
footLen++;
}
do
{
cur = pre[cur];
cycleLen++;
}while(degree[cur] == 1);
if(footLen < cycleLen)
answer = answer * 2 % MOD;
if(footLen > cycleLen)
throw no_solution();
}
}
void countCycle()
{
for(int i = 1; i <= N; i++)
{
if(vis[i])
continue;
int cur = i, len = 0;
do
{
vis[cur] = true;
cur = nxt[cur];
len++;
}while(cur != i);
lenCnt[len]++;
}
}
int main()
{
answer = 1;
ios::sync_with_stdio(false), cin.tie(nullptr);
cin >> N;
for (int i = 1; i <= N; i++)
cin >> nxt[i], degree[nxt[i]]++;
try
{
findCycle();
processCircleBasedTree();
countCycle();
for(int i = 1; i <= N; i++)
{
dp[0] = 1LL;
for(int j = 1; j <= lenCnt[i]; j++)
{
dp[j] = dp[j-1] * (((i % 2 == 0) || i == 1) ? 1 : 2);
if(j >= 2)
dp[j] = (dp[j] + ((((dp[j-2] * (j-1)) % MOD) * i) % MOD)) % MOD;
}
answer = answer * dp[lenCnt[i]] % MOD;
}
cout << answer << endl ;
} catch (const std::exception & x)
{
cout << "0" << endl;
cerr << x.what() << endl;
}
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100100
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define drp(x, a, b) for(int x=a; x>=b; x--)
int st[N<<1], deg[N], d[N], n, vis[N], a[N], tot, num[N], l[N], ans=1, fac[N], inv[N];
typedef long long LL;
int power(int x, int y){
int an = 1;
for(; y; y >>= 1, x = (LL)x * x % mod) if(y & 1) an = (LL)an * x % mod;
return an;
}
int C(int x, int y){
if(x < 0 || y < 0 || x < y) return 0;
return (LL)fac[x] * inv[y] % mod * inv[x - y] % mod;
}
void renew(int &x, const int y){
x += y;
if(x < 0) x += mod;
if(x >= mod) x -= mod;
}
void init(){
fac[0] = inv[0] = fac[1] = inv[1] = 1;
rep(i, 2, n){
fac[i] = (LL)fac[i - 1] * i % mod;
inv[i] = power(fac[i], mod - 2);
}
}
int getdp(){
int fr = 0;
rep(i, 1, tot) st[i + tot] = st[i];
rep(i, 1, tot) if(l[st[i]]){
fr = i;
break;
}
reverse(st + fr + 1, st + fr + tot);
int a = fr + l[st[fr]] - 1, b = a + 1, c = 1;
rep(i, fr + 1, fr + tot - 1)if(l[st[i]]){
int na = i + l[st[i]] - 1, nb = na + 1, cc = 0;
if(a < i) renew(cc, c);
if(b < i) renew(cc, c);
a = na;
b = nb;
c = cc;
}
int ans = 0;
if(a < fr + tot) renew(ans, c);
if(b < fr + tot) renew(ans, c);
return ans;
}
int que[N], h, t;
int main(){
scanf("%d", &n);
rep(i, 1, n)
{
scanf("%d", a+i); deg[a[i]]++;
}
rep(i, 1, n) if(deg[i]>2) return puts("0"), 0;
rep(i, 1, n) d[i] = deg[i];
rep(i, 1, n) if(!d[i]) que[++t] = i;
rep(h, 1, t){
int u = que[h];
--d[a[u]];
vis[u] = 1;
if(!d[a[u]]) que[++t] = a[u];
}
rep(i, 1, n) if(vis[i] && deg[i] >= 2){
puts("0");
exit(0);
}
rep(i, 1, n) if(vis[i] && !deg[i]){
int x = i, cnt = 0;
for(x = i; vis[x]; x = a[x])cnt++;
l[x] = cnt;
}
init();
ans=1;
rep(i, 1, n)if(!vis[i])
{
int u = i, cu = 1;
tot = 0;
int res = 0;
do{
st[++tot] = u;
cu &= !l[u];
vis[u] = 1;
u = a[u];
}while(u != i);
if(cu){
num[tot] ++;
res = 1;
}else{
res = getdp();
}
ans = (LL)ans * res % mod;
}
rep(i, 1, n) if(num[i])
{
int tot = num[i];
int res = 0;
for(int x = 0; x <= tot; x += 2){
int cb = 1;
cb = (LL)cb * power(i, (x >> 1)) % mod;
cb = (LL)cb * C(tot, x) % mod;
cb = (LL)cb * fac[x] % mod;
cb = (LL)cb * inv[x >> 1] % mod;
cb = (LL)cb * power((mod + 1) >> 1, x >> 1) % mod;
if((i > 1) && (i & 1)){
cb = (LL)cb * power(2, tot - x) % mod;
}
renew(res, cb);
}
ans = (LL)ans * res % mod;
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100100
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define drp(x, a, b) for(int x=a; x>=b; x--)
int deg[N], d[N], n, vis[N], a[N], tot, num[N], l[N], ans=1, fac[N], inv[N];
vector<pair<int, int> > fu;
queue<int> q;
int power(int a, int k){
int ret=1;
while(k)
{
if(k&1) ret=1ll*ret*a%mod;
a=1ll*a*a%mod; k>>=1;
}
return ret;
}
int C(int n, int m){
return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int pr(int n){
return 1ll*fac[2*n]*inv[n]%mod*power((mod+1)>>1, n)%mod;
}
void init(){
fac[0]=1;
rep(i, 1, n) fac[i]=1ll*fac[i-1]*i%mod;
inv[n]=power(fac[n], mod-2);
drp(i, n, 1) inv[i-1]=1ll*inv[i]*i%mod;
}
int st[N<<1];
void renew(int &x, const int y){
x += y;
if(x < 0) x += mod;
if(x >= mod) x -= mod;
}
int getdp(){
int fr = 0;
rep(i, 1, tot) st[i + tot] = st[i];
rep(i, 1, tot) if(l[st[i]]){
fr = i;
break;
}
reverse(st + fr + 1, st + fr + tot);
int a = fr + l[st[fr]] - 1, b = a + 1, c = 1;
rep(i, fr + 1, fr + tot - 1)if(l[st[i]]){
int na = i + l[st[i]] - 1, nb = na + 1, cc = 0;
if(a < i) renew(cc, c);
if(b < i) renew(cc, c);
a = na;
b = nb;
c = cc;
}
int ans = 0;
if(a < fr + tot) renew(ans, c);
if(b < fr + tot) renew(ans, c);
return ans;
}
int main(){
scanf("%d", &n);
init();
rep(i, 1, n)
{
scanf("%d", a+i);
deg[a[i]]++;
}
rep(i, 1, n) if(deg[i]>2) return puts("0"), 0;
rep(i, 1, n) d[i]=deg[i];
rep(i, 1, n) if(!d[i]) q.push(i);
while(!q.empty())
{
int u=q.front(); q.pop();
vis[u]=1;
if(--d[a[u]]==0) q.push(a[u]);
}
rep(i, 1, n) if(vis[i] && deg[i]>1) return puts("0"), 0;
rep(i, 1, n) if(vis[i] && !deg[i])
{
int x=i, cnt=0;
for(; vis[x]; x=a[x]) cnt++;
l[x]=cnt;
}
rep(i, 1, n) if(!vis[i])
{
int x=i, fl=0, ass=1; tot=0;
for(; !vis[x]; x=a[x]) st[++tot]=x, fl|=bool(l[x]), vis[x]=1;
if(!fl) num[tot]++;
else
{
fu.clear();
//if(l[i]) fu.pb(mp(l[i], 0));
//for(int p=1, x=a[i]; x!=i; x=a[x]) if(l[x]) fu.pb(mp(l[x], p));
rep(p, 1, tot) if(l[st[p]]) fu.pb(mp(l[st[p]], p));
fu.pb(mp(fu[0].fi, fu[0].se+tot));
int sz=fu.size();
rep(j, 1, sz-1)
{
if(fu[j].fi>fu[j].se-fu[j-1].se) ass=0;
if(fu[j].fi<fu[j].se-fu[j-1].se) ass=2ll*ass%mod;
}
// ass=getdp();
}
ans=1ll*ans*ass%mod;
}
rep(i, 1, n) if(num[i])
{
int m=num[i], ass=0;
rep(j, 0, m/2)
{
int tmp=1ll*C(m, 2*j)*pr(j)%mod*power(i, j)%mod;
if(i>1 && (i&1)) tmp=1ll*tmp*power(2, m-2*j)%mod;
ass=(ass+tmp)%mod;
}
ans=1ll*ans*ass%mod;
}
printf("%d\n", ans);
} | CPP |
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