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For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader r = new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(r);
String num = br.readLine();
int N = Integer.parseInt(num);
String result = "";
if(N < 1000 && N > 0 && N == 1){
result = "AC";
} else {
result = "WA";
}
System.out.println(result);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: N=int(input())
if N==1:
print("AC")
else:
print("WA"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: //Author: Xzirium
//Time and Date: 03:04:29 27 December 2021
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(N);
if(N==1)
{
cout<<"AC"<<endl;
}
else
{
cout<<"WA"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create 6 variables with names<code>varA, varB, varC, varD, varE, varF</code>
Store 6 different values of your choice in the variables such that the variables represent the given primitive types in the given order:
<code>varA</code>: Number
<code>varB</code>: String
<code>varC</code>: Boolean
<code>varD</code>: Undefined
<code>varE</code>: Object
<code>varF</code>: SymbolNo input is required, you just need to create the variables..No output is required for this question..console.log(typeof varA); //prints number
console.log(typeof varB); //prints stirng
console.log(typeof varC); //prints boolean
console.log(typeof varD); //prints undefined
console.log(typeof varE); //prints object
console.log(typeof varF); //prints symbol
, I have written this Solution Code: var varA = 69;
var varB = "hello";
var varC = true;
var varD;
var varE = {};
var varF = Symbol();, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. You have to print the maximum in every <b>K-sized subarray</b> from the left to the right in the array A.
More formally, you have to print N - K + 1 integers X<sub>1</sub>, X<sub>2</sub>, ..., X<sub>N-K+1</sub> such that X<sub>i</sub> (1 <= i <= N - K + 1) is the maximum element in the subarray A<sub>i</sub>, A<sub>i+1</sub>, ..., A<sub>i+K-1</sub>.1. The first line contains an integer N, denoting the size of the array.
2. The second line has N space-separated integers of the array A.
3. The third line contains integer K, denoting the size of the sliding window
<b>Constraints :</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>4</sup> ≤ A[i] ≤ 10<sup>4</sup>
1 ≤ K ≤ NPrint the max of K numbers for each position of sliding windowSample Input:-
8
1 3 -1 -3 5 3 6 7
3
Sample Output:-
3 3 5 5 6 7
Explanation:-
Window position Max
- - - -
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Sample Input:-
1
1
1
Sample Output:-
1 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
int n=Integer.parseInt(bu.readLine());
String s[]=bu.readLine().split(" ");
int a[]=new int[n],i;
PriorityQueue<int[]> pq=new PriorityQueue<>(new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
if(o1[0]<o2[0]) return 1;
else return -1;
}});
for(i=0;i<n;i++) a[i]=Integer.parseInt(s[i]);
int k=Integer.parseInt(bu.readLine());
for(i=0;i<k;i++) pq.add(new int[]{a[i],i});
sb.append(pq.peek()[0]+" ");
for(i=k;i<n;i++)
{
pq.add(new int[]{a[i],i});
while(!pq.isEmpty() && pq.peek()[1]<=i-k) pq.poll();
sb.append(pq.peek()[0]+" ");
}
System.out.println(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. You have to print the maximum in every <b>K-sized subarray</b> from the left to the right in the array A.
More formally, you have to print N - K + 1 integers X<sub>1</sub>, X<sub>2</sub>, ..., X<sub>N-K+1</sub> such that X<sub>i</sub> (1 <= i <= N - K + 1) is the maximum element in the subarray A<sub>i</sub>, A<sub>i+1</sub>, ..., A<sub>i+K-1</sub>.1. The first line contains an integer N, denoting the size of the array.
2. The second line has N space-separated integers of the array A.
3. The third line contains integer K, denoting the size of the sliding window
<b>Constraints :</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>4</sup> ≤ A[i] ≤ 10<sup>4</sup>
1 ≤ K ≤ NPrint the max of K numbers for each position of sliding windowSample Input:-
8
1 3 -1 -3 5 3 6 7
3
Sample Output:-
3 3 5 5 6 7
Explanation:-
Window position Max
- - - -
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Sample Input:-
1
1
1
Sample Output:-
1 , I have written this Solution Code: N = int(input())
A = [int(x) for x in input().split()]
K = int(input())
def print_max(a, n, k):
max_upto = [0 for i in range(n)]
s = []
s.append(0)
for i in range(1, n):
while (len(s) > 0 and a[s[-1]] < a[i]):
max_upto[s[-1]] = i - 1
del s[-1]
s.append(i)
while (len(s) > 0):
max_upto[s[-1]] = n - 1
del s[-1]
j = 0
for i in range(n - k + 1):
while (j < i or max_upto[j] < i + k - 1):
j += 1
print(a[j], end=" ")
print()
print_max(A, N, K), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. You have to print the maximum in every <b>K-sized subarray</b> from the left to the right in the array A.
More formally, you have to print N - K + 1 integers X<sub>1</sub>, X<sub>2</sub>, ..., X<sub>N-K+1</sub> such that X<sub>i</sub> (1 <= i <= N - K + 1) is the maximum element in the subarray A<sub>i</sub>, A<sub>i+1</sub>, ..., A<sub>i+K-1</sub>.1. The first line contains an integer N, denoting the size of the array.
2. The second line has N space-separated integers of the array A.
3. The third line contains integer K, denoting the size of the sliding window
<b>Constraints :</b>
1 ≤ N ≤ 10<sup>5</sup>
-10<sup>4</sup> ≤ A[i] ≤ 10<sup>4</sup>
1 ≤ K ≤ NPrint the max of K numbers for each position of sliding windowSample Input:-
8
1 3 -1 -3 5 3 6 7
3
Sample Output:-
3 3 5 5 6 7
Explanation:-
Window position Max
- - - -
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Sample Input:-
1
1
1
Sample Output:-
1 , I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
vector<int> maxSwindow(vector<int>& arr, int k) {
vector<int> result;
deque<int> Q(k); // store the indices
// process the first window
for(int i = 0; i < k; i++) {
while(!Q.empty() and arr[i] >= arr[Q.back()])
Q.pop_back();
Q.push_back(i);
}
// process the remaining elements
for(int i = k; i < arr.size(); i++) {
// add the max of the current window
result.push_back(arr[Q.front()]);
// remove the elements going out of the window
while(!Q.empty() and Q.front() <= i - k)
Q.pop_front();
// remove the useless elements
while(!Q.empty() and arr[i] >= arr[Q.back()])
Q.pop_back();
// add the current element in the deque
Q.push_back(i);
}
result.push_back(arr[Q.front()]);
return result;
}
int main()
{
int k, n, m;
cin >> n;
vector<int> nums, res;
for(int i =0; i < n; i++){
cin >> m;
nums.push_back(m);
}
cin >> k;
res = maxSwindow(nums,k);
for(auto i = res.begin(); i!=res.end(); i++)
cout << *i << " ";
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String st = br.readLine();
int len = 0;
int c=0;
for(int i=0;i<st.length();i++){
if(st.charAt(i)=='A'){
c++;
len = Math.max(len,c);
}else{
c=0;
}
}
System.out.println(len);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: S=input()
max=0
flag=0
for i in range(0,len(S)):
if(S[i]=='A' or S[i]=='B'):
if(S[i]=='A'):
flag+=1
if(flag>max):
max=flag
else:
flag=0
print(max), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of characters 'A' or 'B' only, you need to find the maximum length of substring consisting of character 'A' only.The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100000
S consists of characters 'A' or 'B' only.Output a single integer, the answer to the problem.Sample Input
ABAAABBBAA
Sample Output
3
Explanation: Substring from character 3-5 is the longest consisting of As only.
Sample Input
AAAA
Sample Output
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
string s; cin>>s;
int ct = 0;
int ans = 0;
for(char c: s){
if(c == 'A')
ct++;
else
ct=0;
ans = max(ans, ct);
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader(){
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1){
if (c == '\n')break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)c = read();
do{
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)return -ret;return ret;
}
public long nextLong() throws IOException{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.'){
while ((c = read()) >= '0' && c <= '9'){
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)buffer[0] = -1;
}
private byte read() throws IOException{
if (bufferPointer == bytesRead)fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException{
if (din == null)return;
din.close();
}
}
public static void main (String[] args) throws IOException{
Reader sc = new Reader();
int m = sc.nextInt();
int n = sc.nextInt();
int[][] arr = new int[m][n];
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
arr[i][j] = sc.nextInt();
}
}
int max_row_index = 0;
int j = n - 1;
for (int i = 0; i < m; i++) {
while (j >= 0 && arr[i][j] == 1) {
j = j - 1;
max_row_index = i;
}
}
System.out.println(max_row_index);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: r, c = list(map(int, input().split()))
max_count = 0
max_r = 0
for i in range(r):
count = input().count("1")
if count > max_count:
max_count = count
max_r = i
print(max_r), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int a[max1][max1];
signed main()
{
int n,m;
cin>>n>>m;
FOR(i,n){
FOR(j,m){cin>>a[i][j];}}
int cnt=0;
int ans=0;
int res=0;
FOR(i,n){
cnt=0;
FOR(j,m){
if(a[i][j]==1){
cnt++;
}}
if(cnt>res){
res=cnt;
ans=i;
}
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function max1Row(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let j, max_row_index = 0;
j = m - 1;
for (let i = 0; i < n; i++)
{
// Move left until a 0 is found
let flag = false;
// to check whether a row has more 1's than previous
while (j >= 0 && mat[i][j] == 1)
{
j = j - 1; // Update the index of leftmost 1
// seen so far
flag = true;//present row has more 1's than previous
}
// if the present row has more 1's than previous
if (flag)
{
max_row_index = i; // Update max_row_index
}
}
if (max_row_index == 0 && mat[0][m - 1] == 0)
return -1;
return max_row_index;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that takes the integer n as a parameter.
</b>Constraints:</b>
1 <= N <= 100
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print all the even numbers from 1 to n.Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: public static void For_Loop(int n){
for(int i=2;i<=n;i+=2){
System.out.print(i+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves problems on permutation but this time she stuck on a problem and asks for your help.
Given a permutation of N integers as Arr[], your task is to check for each K(1 <= K <= N) there exists a subarray of size K such that it is also a permutation of K integers.
Note:- A permutation of N integers is a sequence of size N where every element from 1- N are present.The first line of input contains a single integer N denoting the size of permutation, the next line of input contains N space separated integers depicting the permutaiton.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers either 1 or 0. Print 1 if there exist a permutation K for the ith number else print 0.Sample Input:-
6
4 5 1 3 2 6
Sample Output:-
1 0 1 0 1 1
Explanation:-
for k=1 permutaion exist from [3, 3]
for k=2 no permutaion exists
for k=3 permutaion exist from [3, 5]
for k=4 no permutaion exists
for k=5 permutaion exist from [1, 5]
for k=6 permutaion exist from [1, 6]
Sample Input:-
6
6 5 4 3 2 1
Sample Output:-
1 1 1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String str1 = br.readLine();
String[] str2 = str1.split(" ");
int[] arr = new int[n];
for(int i = 0; i < n; ++i) {
arr[i] = Integer.parseInt(str2[i]);
}
PermuteTheArray(arr, n);
}
static void PermuteTheArray(int A[], int n)
{
int []arr = new int[n];
for(int i = 0; i < n; i++)
{
arr[A[i] - 1] = i;
}
int mini = n, maxi = 0;
for(int i = 0; i < n; i++)
{
mini = Math.min(mini, arr[i]);
maxi = Math.max(maxi, arr[i]);
if (maxi - mini == i)
System.out.print(1+" ");
else
System.out.print(0+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves problems on permutation but this time she stuck on a problem and asks for your help.
Given a permutation of N integers as Arr[], your task is to check for each K(1 <= K <= N) there exists a subarray of size K such that it is also a permutation of K integers.
Note:- A permutation of N integers is a sequence of size N where every element from 1- N are present.The first line of input contains a single integer N denoting the size of permutation, the next line of input contains N space separated integers depicting the permutaiton.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers either 1 or 0. Print 1 if there exist a permutation K for the ith number else print 0.Sample Input:-
6
4 5 1 3 2 6
Sample Output:-
1 0 1 0 1 1
Explanation:-
for k=1 permutaion exist from [3, 3]
for k=2 no permutaion exists
for k=3 permutaion exist from [3, 5]
for k=4 no permutaion exists
for k=5 permutaion exist from [1, 5]
for k=6 permutaion exist from [1, 6]
Sample Input:-
6
6 5 4 3 2 1
Sample Output:-
1 1 1 1 1 1, I have written this Solution Code: def find_permutations(arr):
max_ind = -1
min_ind = 10000000;
n = len(arr)
index_of = {}
for i in range(n):
index_of[arr[i]] = i + 1
for i in range(1, n + 1):
max_ind = max(max_ind, index_of[i])
min_ind = min(min_ind, index_of[i])
if (max_ind - min_ind + 1 == i):
print(1,end = " ")
else:
#print( "i = ",i)
print(0,end = " ")
n = int(input())
arr = list(map(int, input().split()))
find_permutations(arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves problems on permutation but this time she stuck on a problem and asks for your help.
Given a permutation of N integers as Arr[], your task is to check for each K(1 <= K <= N) there exists a subarray of size K such that it is also a permutation of K integers.
Note:- A permutation of N integers is a sequence of size N where every element from 1- N are present.The first line of input contains a single integer N denoting the size of permutation, the next line of input contains N space separated integers depicting the permutaiton.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers either 1 or 0. Print 1 if there exist a permutation K for the ith number else print 0.Sample Input:-
6
4 5 1 3 2 6
Sample Output:-
1 0 1 0 1 1
Explanation:-
for k=1 permutaion exist from [3, 3]
for k=2 no permutaion exists
for k=3 permutaion exist from [3, 5]
for k=4 no permutaion exists
for k=5 permutaion exist from [1, 5]
for k=6 permutaion exist from [1, 6]
Sample Input:-
6
6 5 4 3 2 1
Sample Output:-
1 1 1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
pair<int,int> ans[100005];
signed main(){
// freopen("ou.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int t;t=1;
while(t--){
int n;
cin>>n;
int a[n+10];
int b[n+10];
FOR(i,n){
cin>>a[i];
b[a[i]]=i;
}
ans[1].first=b[1];
ans[1].second=b[1];
for(int i=2;i<=n;i++){
ans[i].first=min(ans[i-1].first,b[i]);
ans[i].second=max(ans[i-1].second,b[i]);
}
for(int i=1;i<=n;i++){
if(ans[i].second-ans[i].first+1==i){cout<<1<<" ";}
else{
cout<<0<<" ";
}
}
END;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given an array of several arrays that each contain integers and your goal is to write a function that
will sum up all the numbers in all the arrays. For example, if the input is [[3, 2], [1], [4, 12]] then your
program should output 22 because 3 + 2 + 1 + 4 + 12 = 22An array containing arrays which can contain any number of elements.Sum of all the elements in all of the arrays.Sample input:-
[[3, 2], [1], [4, 12]]
Sample output:-
22
Explanation:-
3 + 2 + 1 + 4 + 12 = 22, I have written this Solution Code: function sum_array(arr) {
// store our final answer
var sum = 0;
// loop through entire array
for (var i = 0; i < arr.length; i++) {
// loop through each inner array
for (var j = 0; j < arr[i].length; j++) {
// add this number to the current final sum
sum += arr[i][j];
}
}
console.log(sum);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes, your task is to traverse the list and print its elements.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>printList()</b> that takes head node of the linked list as parameter.
<b>Constraints:</b>
1 <=N <= 1000
1 <=Node.data<= 1000For each testcase you need to print the elements of the linked list separated by space. The driver code will take care of printing new line.Sample input
5
2 4 5 6 7
Sample output
2 4 5 6 7
Sample Input
4
1 2 3 5
Sample output
1 2 3 5, I have written this Solution Code: public static void printList(Node head) {
while(head!=null)
{
System.out.print(head.val + " ");
head=head.next;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args) throws IOException {
Reader sc=new Reader();
int n=sc.nextInt();
int[] a=new int[n];
int counter=0;
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
if(a[i]%2!=0) counter++;
}
String ans="NO";
if((a[0]%2!=0)&&(a[n-1]%2!=0)&&(n%2!=0))
ans="YES";
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code:
def fun(n,arr):
if n%2==0:
if arr[0]%2==1 and arr[n-1]%2==1:
for i in range(1,n):
if arr[i]%2==1 and arr[i-1]%2==1:
return "YES"
else:
return "NO"
else:
if arr[0]%2==1 and arr[n-1]%2==1:
return "YES"
else:
return "NO"
n=int(input())
arr=[int(i) for i in input().split()]
print(fun(n,arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code: 'use strict'
function main(input)
{
const inputList = input.split('\n');
const arrSize = Number(inputList[0].split(' ')[0])
const arr = inputList[1].split(' ').
map(x => Number(x))
if(arrSize %2 == 0 || Number(arr[0]) %2 == 0 ||
Number(arr[arrSize-1])%2 == 0)
console.log("NO")
else console.log("YES")
}
main(require("fs").readFileSync("/dev/stdin", "utf8"));, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create 6 variables with names<code>varA, varB, varC, varD, varE, varF</code>
Store 6 different values of your choice in the variables such that the variables represent the given primitive types in the given order:
<code>varA</code>: Number
<code>varB</code>: String
<code>varC</code>: Boolean
<code>varD</code>: Undefined
<code>varE</code>: Object
<code>varF</code>: SymbolNo input is required, you just need to create the variables..No output is required for this question..console.log(typeof varA); //prints number
console.log(typeof varB); //prints stirng
console.log(typeof varC); //prints boolean
console.log(typeof varD); //prints undefined
console.log(typeof varE); //prints object
console.log(typeof varF); //prints symbol
, I have written this Solution Code: var varA = 69;
var varB = "hello";
var varC = true;
var varD;
var varE = {};
var varF = Symbol();, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: def factorial(n):
if(n == 1):
return 1
return n * factorial(n-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: static int Factorial(int N)
{
if(N==0){
return 1;}
return N*Factorial(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*1).<b>User Task</b>
Since this is a functional problem, you don't have to worry about the input. You just have to complete the function <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: // n is the input number
function factorial(n) {
// write code here
// do not console.log
// return the answer as a number
if (n == 1 ) return 1;
return n * factorial(n-1)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of N integers A1, A2,. . An. You have to calculate prosum of the array.
Prosum is defined as the sum of all Ai x Aj over all pairs (i, j) such that 1 <= i < j <= N, modulo (10^9 + 7)The first line of the input contains a single integer N
The next line of the input contains N different integers A1, A2,. . An
Constraints:
1) 2 <= N <= 2 x 10^5
2) 0 <= Ai <= 10^9Print the answerSample Input 1:
4
1 2 3 4
Sample Output 1:
35
Explanation:
1 x 2 + 1 x 3 + 1 x 4 + 2 x 3 + 2 x 4 + 3 x 4 = 35
Sample Input 2:
4
141421356 17320508 22360679 244949
Sample Output 2:
437235829, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for (int i = a; i < (int)(n); i++)
int main() {
long long N, M=1000000007, Sum=0, Ans=0;
cin >> N;
long long A[N];
rep(i, 0, N){
cin >> A[i];
Sum=(Sum+A[i])%M;
}
rep(i, 0, N-1){
Sum-=A[i];
if(Sum<0) Sum+=M;
Ans=(Ans+A[i]*Sum%M)%M;
}
cout << Ans << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(in.readLine());
while(t-- > 0){
String str[] = in.readLine().trim().split(" ");
int k = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int z = Integer.parseInt(str[2]);
int cnt = Math.abs(x-z);
int rem = k - cnt;
if(cnt < rem){
if(cnt != 0){
System.out.println(cnt-1);
}
else{
System.out.println(0);
}
}
else if(rem < cnt){
if(rem != 0){
System.out.println(rem-1);
}
else{
System.out.println(0);
}
}
else{
System.out.println(0);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: k = int(input())
for i in range(k):
z = 0
a = input().split()
a = [int(i) for i in a]
rad = 360/a[0]
dia = abs(a[2]-a[1])
ang = rad*dia
if(ang/2 > 90):
z = a[0]- dia - 1
elif(ang/2 < 90):
z = dia - 1
else:
z = 0
print(z)
a.clear(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int t,k,r,s,a,b;
cin>>t;
while(t--)
{
cin>>k>>r>>s;
b=max(r,s);
a=min(r,s);
if(k%2==0&&(b-a)==k/2)
cout<<0;
else
if((k%2!=0&&(b-a)<=k/2)||(k%2==0&&(b-a)<k/2))
cout<<b-a-1;
else if((b-a)>k/2)
cout<<k-(b-a)-1;
cout<<endl;
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N * M. Find the maximum path sum in matrix. The maximum path sum is sum of path elements from first row to last row where you are allowed to move only down(from (i,j) to (i+1,j)) or diagonally to left(from (i,j) to (i+1,j-1)) or right(from (i,j) to (i+1,j+1)). You can start from any element in first row. You can finish on any element of the last row.First line contains T, the number of test case.
For each test case you will be given two integers, N and M, denoting dimensions of the matrix
Each of the next N lines contains M integers, A[i][j] denoting the value of element at cell (i, j) of the matrix.
1 <= T <= 100
1 <= N, M <= 100
1 <= A[i][j] <= 1000For each test case, print a single line containing the maximum path sum as described in the problem statementSample input 1:
2
1 2
4 8
4 3
1 8 5
5 3 4
6 2 3
4 1 7
Sample Output 1:
8
23
Explanation:
For 2nd test case, max path sum is of the following path: (1, 2) -> (2, 1) -> (3, 1) -> (4, 1), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
int[][] dp = new int[102][102];
while(t-- > 0) {
String[] s = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int m = Integer.parseInt(s[1]);
int ans = 0;
int k = 0;
for(int i = 1; i <= n; i++) {
String[] temp = br.readLine().split(" ");
for(int j = 1; j <= m; j++) {
int val = Integer.parseInt(temp[j-1]);
dp[i][j] = val + Math.max(
Math.max(
dp[i-1][j-1],
dp[i-1][j]
),
dp[i-1][j+1]
);
if(i == n) {
ans = Math.max(ans, dp[i][j]);
}
}
k = k^1;
}
System.out.println(ans);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N * M. Find the maximum path sum in matrix. The maximum path sum is sum of path elements from first row to last row where you are allowed to move only down(from (i,j) to (i+1,j)) or diagonally to left(from (i,j) to (i+1,j-1)) or right(from (i,j) to (i+1,j+1)). You can start from any element in first row. You can finish on any element of the last row.First line contains T, the number of test case.
For each test case you will be given two integers, N and M, denoting dimensions of the matrix
Each of the next N lines contains M integers, A[i][j] denoting the value of element at cell (i, j) of the matrix.
1 <= T <= 100
1 <= N, M <= 100
1 <= A[i][j] <= 1000For each test case, print a single line containing the maximum path sum as described in the problem statementSample input 1:
2
1 2
4 8
4 3
1 8 5
5 3 4
6 2 3
4 1 7
Sample Output 1:
8
23
Explanation:
For 2nd test case, max path sum is of the following path: (1, 2) -> (2, 1) -> (3, 1) -> (4, 1), I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e18 + 9;
int a[N], dp[N][N];
void solve(){
int n, m, ans = 0; cin >> n >> m;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
int p; cin >> p;
dp[i][j] = p + max({dp[i-1][j-1], dp[i-1][j], dp[i-1][j+1]});
if(i == n)
ans = max(ans, dp[i][j]);
}
}
cout << ans << endl;
}
void testcases(){
int tt = 1;
cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code:
L,B,H=input().split()
a=4*(int(L)+int(B)+int(H))
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code: static int Perimeter(int L, int B, int H){
return 4*(L+B+H);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: static int LikesBoth(int N, int A, int B){
return (A+B-N);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: def LikesBoth(N,A,B):
return (A+B-N)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: int LikesBoth(int N,int A, int B){
return (A+B-N);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: int LikesBoth(int N,int A, int B){
return (A+B-N);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton being a student of science gifted an <b>nXn</b> matrix to his girlfriend on her birthday and challenge her to solve the matrix. The matrix consists of natural numbers up to 500(1 ≤ n ≤ 500). Newton instructed a few things about matrix to his girlfriend. He said there are two types of symbols on the matrix( L and R ), you need to rotate the matrix 90 degrees to the left for every L symbol and rotate the matrix 90 degrees right for every R symbol. The symbols were only 3 characters in length.
Help Newton's girlfriend to display the final matrix at the end of the complete turns.The first line consists of one integer n - the size of the matrix.
In the next n lines, you are given n integers, numbers can range from 1 to 500.
<b>Constraints<b>
1 ≤ n ≤ 500Output the final matrix n X n.
<b>Note</b> You should not print any whitespace or newline if it is not necessary.Sample input:
2
1 2
3 4
RLR
Sample output:
3 1
4 2, I have written this Solution Code: #include <bits/stdc++.h>
#define ll long long int
using namespace std;
void solve()
{
ll n;
cin >> n;
vector<vector<ll>> a(n, vector<ll>(n, 0));
for (ll i = 0; i < n; i++)
{
for (ll j = 0; j < n; j++)
{
cin >> a[i][j];
}
}
string s;
cin >> s;
ll count = 0;
for (auto i : s)
{
if (i == 'R')
count++;
else
count--;
}
if (count > 0)
{
while (count--)
{
for (ll i = 0; i < n; i++)
{
for (ll j = 0; j < i; j++)
{
swap(a[i][j], a[j][i]);
}
}
for (ll i = 0; i < n; i++)
{
for (ll j = 0; j < n / 2; j++)
{
swap(a[i][j], a[i][n - 1 - j]);
}
}
}
}
else
{
while (count++)
{
for (ll i = 0; i < n; i++)
{
for (ll j = 0; j < i; j++)
{
swap(a[i][j], a[j][i]);
}
}
for (ll j = 0; j < n; j++)
{
for (ll i = 0; i < n / 2; i++)
{
swap(a[i][j], a[n - 1 - i][j]);
}
}
}
}
for (ll i = 0; i < n; i++)
{
for (ll j = 0; j < n; j++)
{
cout << a[i][j] << " ";
}
cout << "\n";
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton being a student of science gifted an <b>nXn</b> matrix to his girlfriend on her birthday and challenge her to solve the matrix. The matrix consists of natural numbers up to 500(1 ≤ n ≤ 500). Newton instructed a few things about matrix to his girlfriend. He said there are two types of symbols on the matrix( L and R ), you need to rotate the matrix 90 degrees to the left for every L symbol and rotate the matrix 90 degrees right for every R symbol. The symbols were only 3 characters in length.
Help Newton's girlfriend to display the final matrix at the end of the complete turns.The first line consists of one integer n - the size of the matrix.
In the next n lines, you are given n integers, numbers can range from 1 to 500.
<b>Constraints<b>
1 ≤ n ≤ 500Output the final matrix n X n.
<b>Note</b> You should not print any whitespace or newline if it is not necessary.Sample input:
2
1 2
3 4
RLR
Sample output:
3 1
4 2, I have written this Solution Code: n = int(input())
matrix = []
for _ in range(n):
matrix.append(input().strip().split())
symbols = list(input().strip())
instructions = sum(1 if i == 'R' else -1 for i in symbols)
if instructions in (1, -3):
rotate_matrix = [['0'] * n for _ in range(n)]
for i in range(n):
for j in range(n):
rotate_matrix[i][j] = matrix[n - j - 1][i]
matrix = rotate_matrix
elif instructions in (-1, 3):
rotate_matrix = [['0'] * n for _ in range(n)]
for i in range(n):
for j in range(n):
rotate_matrix[i][j] = matrix[j][n - i - 1]
matrix = rotate_matrix
for i in range(n):
print(' '.join(matrix[i])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
int max = 1000001;
boolean isNotPrime[] = new boolean[max];
ArrayList<Integer> arr = new ArrayList<Integer>();
isNotPrime[0] = true; isNotPrime[1] = true;
for (int i=2; i*i <max; i++) {
if (!isNotPrime[i]) {
for (int j=i*i; j<max; j+= i) {
isNotPrime[j] = true;
}
}
}
for(int i=2; i<max; i++) {
if(!isNotPrime[i]) {
arr.add(i);
}
}
while(t-- > 0) {
String str[] = br.readLine().trim().split(" ");
int l = Integer.parseInt(str[0]);
int r = Integer.parseInt(str[1]);
System.out.println(primeRangeSum(l,r,arr));
}
}
static long primeRangeSum(int l , int r, ArrayList<Integer> arr) {
long sum = 0;
for(int i=l; i<=r;i++) {
sum += arr.get(i-1);
}
return sum;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
pri = []
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
pri.append(p)
return pri
N = int(input())
X = []
prim = SieveOfEratosthenes(1000000)
for i in range(1,len(prim)):
prim[i] = prim[i]+prim[i-1]
for i in range(N):
nnn = input()
X.append((int(nnn.split()[0]),int(nnn.split()[1])))
for xx,yy in X:
if xx==1:
print(prim[yy-1])
else:
print(prim[yy-1]-prim[xx-2])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
vector<int> v;
v.push_back(0);
for(int i = 2; i < N; i++){
if(a[i]) continue;
v.push_back(i);
for(int j = i*i; j < N; j += i)
a[j] = 1;
}
int p = 0;
for(auto &i: v){
i += p;
p = i;
}
int t; cin >> t;
while(t--){
int l, r;
cin >> l >> r;
cout << v[r] - v[l-1] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
if(n==1){
System.out.println(2);
}else{
int after = afterPrime(n);
int before = beforePrime(n);
if(before>after){
System.out.println(n+after);
}
else{System.out.println(n-before);}
}
}
public static boolean isPrime(int n)
{
int count=0;
for(int i=2;i*i<n;i++)
{
if(n%i==0)
count++;
}
if(count==0)
return true;
else
return false;
}
public static int beforePrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n-1;
c++;
}
}
}
public static int afterPrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n+1;
c++;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: from math import sqrt
def NearPrime(N):
if N >1:
for i in range(2,int(sqrt(N))+1):
if N%i ==0:
return False
break
else: return True
else: return False
N=int(input())
i =0
while NearPrime(N-i)==False and NearPrime(N+i)==False:
i+=1
if NearPrime(N-i) and NearPrime(N+i):print(N-i)
elif NearPrime(N-i):print(N-i)
elif NearPrime(N+i): print(N+i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
bool isPrime(int n){
if(n<=1)
return false;
for(int i=2;i*i<=n;++i)
if(n%i==0)
return false;
return true;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n==1)
cout<<"2";
else{
int v1=n,v2=n;
while(isPrime(v1)==false)
--v1;
while(isPrime(v2)==false)
++v2;
if(v2-n==n-v1)
cout<<v1;
else{
if(v2-n<n-v1)
cout<<v2;
else
cout<<v1;
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Subtract two numbers without using arithmetic operators and print the result.Only line contains two integers a and b.
<b>Constraints</b>
1 <= a, b <= 10<sup>9</sup>print the result of (a- b).Input:
3 7
Output:
-4, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int x,y;
x=Integer.parseInt(in.next());
y=Integer.parseInt(in.next());
while (y != 0){
int borrow = (~x) & y;
x = x ^ y;
y = borrow << 1;
}
out.print(x);
out.close();
}
static class InputReader {
BufferedReader reader;
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());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
System.out.print(nonRepeatChar(s));
}
static int nonRepeatChar(String s){
char count[] = new char[256];
for(int i=0; i< s.length(); i++){
count[s.charAt(i)]++;
}
for (int i=0; i<s.length(); i++) {
if (count[s.charAt(i)]==1){
return i;
}
}
return -1;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: from collections import defaultdict
s=input()
d=defaultdict(int)
for i in s:
d[i]+=1
ans=-1
for i in range(len(s)):
if(d[s[i]]==1):
ans=i
break
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-10 12:51:16
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int firstUniqChar(string s) {
map<char, int> charCount;
int len = s.length();
for (int i = 0; i < len; i++) {
charCount[s[i]]++;
}
for (int i = 0; i < len; i++) {
if (charCount[s[i]] == 1)
return i;
}
return -1;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
string str;
cin>>str;
cout<<firstUniqChar(str)<<"\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A graph is given with N nodes and M edges. The first visit to a vertex i will give 2<sup>i-1</sup> -1 score. Initially, we are at node 1. Then calculate the maximum score that we can get if we are only allowed to move to an adjacent vertex at most K times.First line contains 3 integers separated by space denoting N (1<=N<=60) , M (1<=M<=N*(N-1)/2) , and K (1<=K<=15), the number of nodes, edges, and the maximum move allowed respectively.
Then, M lines follow with each line containing two integers u and v representing edge between vertex u and v (u≠v).Output maximum score in first line.Input:
4 3 1
1 2
1 3
1 4
Output:
7
Explanation:
Alice can move to node 2 or node 3 or node 4. Here node 4 has (2^3 - 1 = 7)coins., I have written this Solution Code: //*** Author: ShlokG ***//
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
typedef long long int ll;
typedef unsigned long long int ull;
#define pb push_back
#define endl "\n"
const ll MAX=1e18+5;
ll mod=1e9+7;
ll mod1=998244353;
const int infi = 1e9;
const ll INF=1e18;
void precompute(){
}
void TEST_CASE(){
int n,m,k;
cin >> n >> m >> k;
int mat[100][100];
int dp[1<<17][17];
vector<int> v;
for(int i=0 ; i<100 ; i++) for(int j=0 ; j<100 ; j++) mat[i][j]=1000;
for(int i=0 ; i<100 ; i++) mat[i][i]=0;
for(int i=0 ; i<m ; i++){
int x,y;
cin >> x >> y;
mat[x-1][y-1]=mat[y-1][x-1]=1;
}
for(int i=0 ; i<n ; i++) for(int j=0 ; j<n ; j++) for(int l=0 ; l<n ; l++) mat[j][l] = min (mat [j] [l], mat [j] [i] + mat [i] [l ]);
v.push_back (0);
ll ret = 0 ;
for(int i=n-1 ; i>0 ; i--) {
if(v.size()>=k+1) break;
v.push_back(i);
memset(dp,0x3f,sizeof(dp));
dp[1][0]=0;
for(int mask=1 ; mask<(1<<v.size()) ; mask++){
for(int x=0 ; x<v.size() ; x++) if(mask&(1<<x)) {
for(int y=0 ; y<v.size() ; y++) if(x!=y && (mask&(1<<y))) dp[mask][x] = min(dp[mask][x], dp[mask^(1<<x)][y] + mat[v[x]][v[y]]);
}
}
int ok = 0 ;
for(int j=0 ; j<v.size() ; j++) if(dp[(1<<v.size())-1][j]<=k) ok++;
if (ok == 0 ) v.pop_back();
else ret += ( 1LL << i) -1 ;
}
cout << ret << endl;
}
signed main(){
fast;
//freopen ("INPUT.txt","r",stdin);
//freopen ("OUTPUT.txt","w",stdout);
int test=1,TEST=1;
precompute();
//cin >> test;
while(test--){
//cout << "Case #" << TEST++ << ": ";
TEST_CASE();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2*N matrix in which each cell contains some candies in it. You are at the top left corner of the matrix and want to reach the bottom right corner of the matrix i. e from (1, 1) to (2, N). You can only move right or down. You have to find the maximum number of candies you can collect in your journey.The first line of input contains a single integer N. The second line of input contains N spaces separated integers. The last line of input contains N space-separated integers.
<b>Constraints:-</b>
2 <= N <= 10000
1 <= Matrix[i][j] <= 100000Print the maximum amount of candies you can have at the end of your journey.Sample Input 1:-
5
1 3 5 6 2
2 4 8 4 2
Sample Output 1:-
23
Sample Input 2:-
4
1 1 1 1
1 1 1 1
Sample Output 2:-
5, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n][2];
for(int i=0;i<n;i++){
cin>>a[i][0];
}
for(int i=0;i<n;i++){
cin>>a[i][1];
}
for(int i=1;i<n;i++){
a[i][0]+=a[i-1][0];
}
for(int i = n-2;i>=0;i--){
a[i][1]+=a[i+1][1];
}
int ans=0;
for(int i=0;i<n;i++){
ans=max(a[i][0]+a[i][1],ans);
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code:
L,B,H=input().split()
a=4*(int(L)+int(B)+int(H))
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code: static int Perimeter(int L, int B, int H){
return 4*(L+B+H);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException
{
StringBuilder out=new StringBuilder();
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int test=Integer.parseInt(br.readLine());
while(test-->0)
{
int n=Integer.parseInt(br.readLine());
String s=br.readLine();
int c1=0;
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='1') c1++;
}
if(c1%4==0) out.append("1\n");
else if(c1==s.length() && (c1/2)%2==0) out.append("1\n");
else
out.append("0\n");
}
System.out.print(out);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: T=int(input())
for i in range(T):
n=int(input())
a=input()
count_1=0
for i in a:
if i=='1':
count_1+=1
if count_1%2==0 and ((count_1)//2)%2==0:
print('1')
else:
print('0'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main() {
int t;
cin >> t;
for(int i=0; i<t; i++) {
int n;
cin >> n;
string s;
cin >> s;
int cnt = 0;
for(int j=0; j<n; j++) {
if(s[j] == '1') cnt++;
}
if(cnt % 4 == 0) cout << 1 << "\n";
else cout << 0 << "\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader sc= new BufferedReader(new InputStreamReader(System.in));
int a=0, b=0, c=0, d=0;
long z=0;
String[] str;
str = sc.readLine().split(" ");
a= Integer.parseInt(str[0]);
b= Integer.parseInt(str[1]);
c= Integer.parseInt(str[2]);
d= Integer.parseInt(str[3]);
BigInteger m = new BigInteger("1000000007");
BigInteger n = new BigInteger("1000000006");
BigInteger zero = new BigInteger("0");
BigInteger ans, y;
if(d==0){
z =1;
}else{
z = (long)Math.pow(c, d);
}
if(b==0){
y= zero;
}else{
y = (BigInteger.valueOf(b)).modPow((BigInteger.valueOf(z)), n);
}
if(y == zero){
System.out.println("1");
}else if(a==0){
System.out.println("0");
}else{
ans = (BigInteger.valueOf(a)).modPow(y, m);
System.out.println(ans);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int powmod(int a, int b, int c = MOD){
int ans = 1;
while(b){
if(b&1){
ans = (ans*a)%c;
}
a = (a*a)%c;
b >>= 1;
}
return ans;
}
void solve(){
int a, b, c, d; cin>>a>>b>>c>>d;
int x = pow(c, d);
int y = powmod(b, x, MOD-1);
int ans = powmod(a, y, MOD);
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: static int equationSum(int n)
{
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: def equationSum(N) :
re=(N*(N+1))//2
re=re*re
re=re+9*((N*(N+1))//2)
re=re+4*N
return re
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String[] str = br.readLine().split(" ");
String ref = str[0];
for(String s: str){
if(s.length()<ref.length()){
ref = s;
}
}
for(int i=0; i<n; i++){
if(str[i].contains(ref) == false){
ref = ref.substring(0, ref.length()-1);
}
}
if(ref.length()>0)
System.out.println(ref);
else
System.out.println(-1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: def longestPrefix( strings ):
size = len(strings)
if (size == 0):
return -1
if (size == 1):
return strings[0]
strings.sort()
end = min(len(strings[0]), len(strings[size - 1]))
i = 0
while (i < end and
strings[0][i] == strings[size - 1][i]):
i += 1
pre = strings[0][0: i]
if len(pre) > 1:
return pre
else:
return -1
N=int(input())
strings=input().split()
print( longestPrefix(strings) ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: function commonPrefixUtil(str1,str2)
{
let result = "";
let n1 = str1.length, n2 = str2.length;
// Compare str1 and str2
for (let i = 0, j = 0; i <= n1 - 1 && j <= n2 - 1; i++, j++) {
if (str1[i] != str2[j]) {
break;
}
result += str1[i];
}
return (result);
}
// n is number of individual space seperated strings inside strings variable,
// strings is the string which contains space seperated words.
function longestCommonPrefix(strings,n){
// write code here
// do not console,log answer
// return the answer as string
if(n===1) return strings;
const arr= strings.split(" ")
let prefix = arr[0];
for (let i = 1; i <= n - 1; i++) {
prefix = commonPrefixUtil(prefix, arr[i]);
}
if(!prefix) return -1;
return (prefix);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
string a[n];
int x=INT_MAX;
for(int i=0;i<n;i++){
cin>>a[i];
x=min(x,(int)a[i].length());
}
string ans="";
for(int i=0;i<x;i++){
for(int j=1;j<n;j++){
if(a[j][i]==a[0][i]){continue;}
goto f;
}
ans+=a[0][i];
}
f:;
if(ans==""){cout<<-1;return 0;}
cout<<(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader sc= new BufferedReader(new InputStreamReader(System.in));
int a=0, b=0, c=0, d=0;
long z=0;
String[] str;
str = sc.readLine().split(" ");
a= Integer.parseInt(str[0]);
b= Integer.parseInt(str[1]);
c= Integer.parseInt(str[2]);
d= Integer.parseInt(str[3]);
BigInteger m = new BigInteger("1000000007");
BigInteger n = new BigInteger("1000000006");
BigInteger zero = new BigInteger("0");
BigInteger ans, y;
if(d==0){
z =1;
}else{
z = (long)Math.pow(c, d);
}
if(b==0){
y= zero;
}else{
y = (BigInteger.valueOf(b)).modPow((BigInteger.valueOf(z)), n);
}
if(y == zero){
System.out.println("1");
}else if(a==0){
System.out.println("0");
}else{
ans = (BigInteger.valueOf(a)).modPow(y, m);
System.out.println(ans);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int powmod(int a, int b, int c = MOD){
int ans = 1;
while(b){
if(b&1){
ans = (ans*a)%c;
}
a = (a*a)%c;
b >>= 1;
}
return ans;
}
void solve(){
int a, b, c, d; cin>>a>>b>>c>>d;
int x = pow(c, d);
int y = powmod(b, x, MOD-1);
int ans = powmod(a, y, MOD);
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
String[] line = br.readLine().split(" ");
int happyBalloons = 0;
for(int i=1;i<=line.length;++i){
int num = Integer.parseInt(line[i-1]);
if(num%2 == i%2 ){
happyBalloons++;
}
}
System.out.println(happyBalloons);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: x=int(input())
arr=input().split()
for i in range(0,x):
arr[i]=int(arr[i])
count=0
for i in range(0,x):
if(arr[i]%2==0 and (i+1)%2==0):
count+=1
elif (arr[i]%2!=0 and (i+1)%2!=0):
count+=1
print (count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N balloons numbered from 1 to N. Every balloon has an another integer value Arr[i] assigned to it where i varies from 1 to N, and i represents the number of balloon.
A balloon at an odd position (i = 1, 3, 5. ) is happy if Arr[i] is odd.
A balloon at an even position (i = 2, 4, 6. ) is happy if Arr[i] is even.
Find the number of happy balloons.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, Arr[1], Arr[2], Arr[3],. , Arr[N].
Constrains
1 <= N <= 200000
1 <= Arr[i] <= 1000000Output a single integer, the number of happy balloons.Sample Input
5
1 3 4 6 7
Sample Output
3
Explanation
Happy balloons are balloons numbered 1, 4, 5.
Sample Input
5
1 2 3 4 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
int ans = 0;
For(i, 1, n+1){
int a; cin>>a;
if(i%2 == a%2)
ans++;
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ketani is fond of super simple problems. This time he has come up with another simple problem. Given two non negative numbers A and B. Find the number of multiples of k in the range [A, B] (all integers between A and B, A and B inclusive).The first and the only line of input contains three integers: A B k.
Constraints
0 <= a <= b <= 1000000000000000000 (10^18)
1 <= x <= 1000000000000000000 (10^18)Output a single integer, the number of multiples of k in the given range.Sample Input
4 8 2
Sample Output
3
Explanation: The divisors of 2 in the range [4, 8] are 4, 6, and 8.
Sample Input:
0 5 1
Sample Output:
6, I have written this Solution Code: a,b,k = list(map(int,input().strip().split()))
if a == 0:
count = 1
a =1
else:
count = 0
if a%k == 0 :
count += b//k - a//k +1
else:
count += b//k - a//k
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ketani is fond of super simple problems. This time he has come up with another simple problem. Given two non negative numbers A and B. Find the number of multiples of k in the range [A, B] (all integers between A and B, A and B inclusive).The first and the only line of input contains three integers: A B k.
Constraints
0 <= a <= b <= 1000000000000000000 (10^18)
1 <= x <= 1000000000000000000 (10^18)Output a single integer, the number of multiples of k in the given range.Sample Input
4 8 2
Sample Output
3
Explanation: The divisors of 2 in the range [4, 8] are 4, 6, and 8.
Sample Input:
0 5 1
Sample Output:
6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String[] inputs = br.readLine().split(" ");
long A = Long.parseLong(inputs[0]);
long B = Long.parseLong(inputs[1]);
long k = Long.parseLong(inputs[2]);
long minMultiple = A/k;
long maxMultiple = B/k;
long compensattion = 0;
if(A % k == 0){
compensattion = 1;
}
long answer = maxMultiple+compensattion-minMultiple;
System.out.println(answer);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ketani is fond of super simple problems. This time he has come up with another simple problem. Given two non negative numbers A and B. Find the number of multiples of k in the range [A, B] (all integers between A and B, A and B inclusive).The first and the only line of input contains three integers: A B k.
Constraints
0 <= a <= b <= 1000000000000000000 (10^18)
1 <= x <= 1000000000000000000 (10^18)Output a single integer, the number of multiples of k in the given range.Sample Input
4 8 2
Sample Output
3
Explanation: The divisors of 2 in the range [4, 8] are 4, 6, and 8.
Sample Input:
0 5 1
Sample Output:
6, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// Driver code
int main()
{
unsigned long long a,b,k,p,q,s=0;
cin>>a>>b>>k;
p=a/k;
if(a%k==0){p--;}
q=b/k;
cout<<max(s,q-p);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader sc= new BufferedReader(new InputStreamReader(System.in));
int a=0, b=0, c=0, d=0;
long z=0;
String[] str;
str = sc.readLine().split(" ");
a= Integer.parseInt(str[0]);
b= Integer.parseInt(str[1]);
c= Integer.parseInt(str[2]);
d= Integer.parseInt(str[3]);
BigInteger m = new BigInteger("1000000007");
BigInteger n = new BigInteger("1000000006");
BigInteger zero = new BigInteger("0");
BigInteger ans, y;
if(d==0){
z =1;
}else{
z = (long)Math.pow(c, d);
}
if(b==0){
y= zero;
}else{
y = (BigInteger.valueOf(b)).modPow((BigInteger.valueOf(z)), n);
}
if(y == zero){
System.out.println("1");
}else if(a==0){
System.out.println("0");
}else{
ans = (BigInteger.valueOf(a)).modPow(y, m);
System.out.println(ans);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four integers a, b, c, and d. Find the value of a<sup>(b<sup>(c<sup>d</sup>)</sup>)</sup> modulo 1000000007.
(Fact: 0<sup>0</sup> = 1)The first and the only line of input contains 4 integers a, b, c, and d.
Constraints
1 <= a, b, c, d <= 12Output a single integer, the answer modulo 1000000007.Sample Input
2 2 2 2
Sample Output
65536
Explanation
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536.
Sample Input
0 7 11 1
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int powmod(int a, int b, int c = MOD){
int ans = 1;
while(b){
if(b&1){
ans = (ans*a)%c;
}
a = (a*a)%c;
b >>= 1;
}
return ans;
}
void solve(){
int a, b, c, d; cin>>a>>b>>c>>d;
int x = pow(c, d);
int y = powmod(b, x, MOD-1);
int ans = powmod(a, y, MOD);
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: def DecimalToBinary(num):
return "{0:b}".format(int(num))
n = int(input())
for i in range(1,n+1):
print(DecimalToBinary(i),end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: import java.util.*;
import java.math.*;
import java.io.*;
class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
StringBuilder s=new StringBuilder("");
for(int i=1;i<=n;i++){
s.append(Integer.toBinaryString(i)+" ");
}
System.out.print(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to generate and print all binary numbers with decimal values from 1 to N.
Try solving this using a queue.The only line of input contains a single integer N.
1 <= N <= 50000Print all binary numbers with decimal values from 1 to N in a single line.Sample Input:
5
Sample Output:
1 10 11 100 101, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 500 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int n; cin >> n;
queue<string> q;
q.push("1");
while(n--){
string x = q.front();
q.pop();
cout << x << " ";
q.push(x + "0");
q.push(x + "1");
}
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: void magicTrick(int a, int b){
cout<<a+b/2;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: static void magicTrick(int a, int b){
System.out.println(a + b/2);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: A,B = map(int,input().split(' '))
C = A+B//2
print(C)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called square if it is some string written twice in a row. For example, the strings "aa", "abcabc", "abab" and "baabaa" are square. But the strings "aaa", "abaaab" and "abcdabc" are not square. For a given string s determine if it is square.The input consists of a string S consisting of lowercase English alphabets.
<b>Constraints</b>
The length of the string is between 1 to 100.Print Yes if the string in the corresponding test case is square, No otherwise.<b>Sample Input 1</b>
aaa
<b>Sample Output 1</b>
No
<b>Sample Input 2</b>
xyxy
<b>Sample Output 2</b>
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t=1;
// cin >> t;
for (int i = 0; i < t; i++){
string s;
cin >> s;
int N = s.size();
if (N % 2 == 1){
cout << "No" << endl;
} else {
if (s.substr(0, N / 2) == s.substr(N / 2)){
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are 3 points, X, Y, and Z on a road.
It takes A hours to travel from X to Y or from Y to X.
It takes B hours to travel from Y to Z or from Z to Y.
It takes C hours to travel from X to Z or from Z to X.
Newton is given the time required to travel between the points, i. e. A, B and C. He wants you to help him to calculate the minimum time needed to visit all the points once, starting from any point of your choice.The first and only line of the input contains 3 integers, A, B, and C.
<b>Constraints:</b>
1 ≤ A, B, C ≤ 1000Output the answer.<b>Sample Input 1:</b>
1 3 4
<b>Sample Output 1:</b>
4
<b>Sample Explanation 1:</b>
The minimum distance to travel all 3 points can be found if he starts at A and then go to B and then to C.
A - > B - > C = 1 + 3 = 4
<b>Sample Input 2:</b>
3 2 3
<b>Sample Output 2:</b>
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) (x).begin(), (x).end()
#define yes "Yes"
#define no "No"
#define ret return
#define xin cin
#define fix(x) fixed << setprecision(x)
#define fore(p, v) for (auto &p : v)
#define mp(a, b) make_pair(a, b)
#define rep(i, l, r) for (ll i = (l); i < (r); i++)
#define inf ((1LL << 62) - (1LL << 31))
int main()
{
ll P, Q, R;
cin >> P >> Q >> R;
ll a = P + Q;
ll b = Q+R;
ll c = R+P;
ll ans = min(a,b);
ans = min(ans, c);
cout << ans << endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The head of the doubly linked list will be given to you, and you must sort it using merge sort.You will be given the head of doubly Linked List and its length.
Constraints:
0<= n <=10000Return the HEAD of the sorted linked list.Sample Input:
5
3 2 1 3 2
Output:
1 2 2 3 3, I have written this Solution Code: class Node:
def __init__(self, data, next=None, prev=None):
self.data = data
self.next = next
self.prev = prev
def push(head, key):
node = Node(key, head)
if head:
head.prev = node
return node
def printDDL(head):
while head:
print(head.data, end=' ')
head = head.next
def split(head):
slow = head
fast = head.next
while fast:
fast = fast.next
if fast:
slow = slow.next
fast = fast.next
return slow
def merge(a, b):
if a is None:
return b
if b is None:
return a
if a.data <= b.data:
a.next = merge(a.next, b)
a.next.prev = a
a.prev = None
return a
else:
b.next = merge(a, b.next)
b.next.prev = b
b.prev = None
return b
def mergesort(head):
if head is None or head.next is None:
return head
a = head
slow = split(head)
b = slow.next
slow.next = None
a = mergesort(a)
b = mergesort(b)
head = merge(a, b)
return head
n=int(input())
keys=list(map(int,input().split()))
head = None
for key in keys:
head = push(head, key)
head = mergesort(head)
printDDL(head), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The head of the doubly linked list will be given to you, and you must sort it using merge sort.You will be given the head of doubly Linked List and its length.
Constraints:
0<= n <=10000Return the HEAD of the sorted linked list.Sample Input:
5
3 2 1 3 2
Output:
1 2 2 3 3, I have written this Solution Code:
Node split(Node head) {
Node fast = head, slow = head;
while (fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
Node temp = slow.next;
slow.next = null;
return temp;
}
Node mergeSort(Node node, int n) {
if (node == null || node.next == null) {
return node;
}
Node second = split(node);
node = mergeSort(node,n);
second = mergeSort(second,n);
return merge(node, second);
}
Node merge(Node first, Node second) {
if (first == null) {
return second;
}
if (second == null) {
return first;
}
if (first.data < second.data) {
first.next = merge(first.next, second);
first.next.prev = first;
first.prev = null;
return first;
} else {
second.next = merge(first, second.next);
second.next.prev = second;
second.prev = null;
return second;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree of N nodes, with root 1 and Q queries with nodes u and v. For each query find the sum of nodes on the shortest path from u to v, where node v is the ancestor of node u.First line contains the integer N and Q, denoting the number of nodes in the binary tree and the number of queries respectively.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
Next Q lines contain two integers u and v
1 <= N <= 10000
1 <= Q <= 100000
1 <= u, v <= NPrint Q lines denoting the sum of nodes on the shortest path from u to vSample Input 1:
6 3
2 4
5 3
-1 -1
-1 -1
6 -1
-1 -1
6 2
3 1
5 5
Sample output 1:
13
6
5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
/
6
Query 1: 6+5+2 = 13
Query 2: 3+2+1 = 6
Query 3: 5 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc= new Scanner(System.in);
int n=sc.nextInt();
int q=sc.nextInt();
int arr[]=new int[n+1];
arr[1]=1;
for(int i=1;i<=n;i++){
int l=sc.nextInt();
int r=sc.nextInt();
if(l!=-1){
arr[l]=l+arr[i];
}
if(r!=-1){
arr[r]=r+arr[i];
}
}
for(int i=0;i<q;i++){
int u=sc.nextInt();
int v=sc.nextInt();
System.out.println(arr[u]-arr[v]+v);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree of N nodes, with root 1 and Q queries with nodes u and v. For each query find the sum of nodes on the shortest path from u to v, where node v is the ancestor of node u.First line contains the integer N and Q, denoting the number of nodes in the binary tree and the number of queries respectively.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
Next Q lines contain two integers u and v
1 <= N <= 10000
1 <= Q <= 100000
1 <= u, v <= NPrint Q lines denoting the sum of nodes on the shortest path from u to vSample Input 1:
6 3
2 4
5 3
-1 -1
-1 -1
6 -1
-1 -1
6 2
3 1
5 5
Sample output 1:
13
6
5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
/
6
Query 1: 6+5+2 = 13
Query 2: 3+2+1 = 6
Query 3: 5 , I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int l[N], r[N], p[N], s[N];
void dfs(int u, int p = 0){
if(u == -1) return;
s[u] = s[p] + u;
dfs(l[u], u);
dfs(r[u], u);
}
signed main() {
IOS;
clock_t start = clock();
int n, q; cin >> n >> q;
for(int i = 1; i <= n; i++){
cin >> l[i] >> r[i];
p[l[i]] = p[r[i]] = i;
}
dfs(1);
while(q--){
int u, v;
cin >> u >> v;
cout << s[u] - s[p[v]] << endl;
}
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(br.readLine());
double arr[] = new double[N];
String str[] = br.readLine().trim().split(" ");
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
double resistance=0;
int equResistance=0;
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
for(int i=0;i<N;i++)
{
resistance=resistance+(1/arr[i]);
}
equResistance = (int)Math.floor((1/resistance));
System.out.println(equResistance);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: r = input("")
r = int(r)
n = input("").split()
resistance=0.0
for i in range(0,r):
resistor = float(n[i])
resistance = resistance + (1/resistor)
print(int(1/resistance)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n; cin >> n;
double s = 0;
for(int i = 1; i <= n; i++){
double p; cin >> p;
s = s + (1/p);
}
s = 1/s;
cout << floor(s);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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