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For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: def profit(C, S):
print(S - C), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: static void Profit(int C, int S){
System.out.println(S-C);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita likes a number if it is stored in an integer while Doraemon likes it when it is stored in a String. Your task is to write a code so that they can easily convert an integer to a string or a string to an integer whenever they want.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the following functions:-
<b>StringToInt()</b> that takes String S as parameter.
<b>IntToString()</b> that takes the integer N as parameter.
Constraints:-
1 <= (Given Number) <= 100Return an integer in <b>StringToInt()</b> while return a integer integer in <b>IntToString()</b>. The driver code will print "<b>Nice Job</b>" if your code is correct otherwise "<b>Wrong answer</b>".Sample Input:-
5
Sample Output:-
Nice Job
Sample Input:-
12
Sample Output:-
Nice Job, I have written this Solution Code: def StringToInt(a):
return int(a)
def IntToString(a):
return str(a)
if __name__ == "__main__":
n = input()
s = StringToInt(n)
if n == str(s):
a=1
# print("Nice Job")
else:
print("Wrong answer")
quit()
p = IntToString(s)
if s == int(p):
print("Nice Job")
else:
print("Wrong answer"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita likes a number if it is stored in an integer while Doraemon likes it when it is stored in a String. Your task is to write a code so that they can easily convert an integer to a string or a string to an integer whenever they want.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the following functions:-
<b>StringToInt()</b> that takes String S as parameter.
<b>IntToString()</b> that takes the integer N as parameter.
Constraints:-
1 <= (Given Number) <= 100Return an integer in <b>StringToInt()</b> while return a integer integer in <b>IntToString()</b>. The driver code will print "<b>Nice Job</b>" if your code is correct otherwise "<b>Wrong answer</b>".Sample Input:-
5
Sample Output:-
Nice Job
Sample Input:-
12
Sample Output:-
Nice Job, I have written this Solution Code: static int StringToInt(String S)
{
return Integer.parseInt(S);
}
static String IntToString(int N){
return String.valueOf(N);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tom is currently playing in the park which is in the form of NxM grid. He is currently at (0,0) and wants to reach (N-1, M-1). If Tom is currently at (a,b), then after a move Tom can be at (a-1,b) or (a+1,b) or(a,b-1) or (a,b+1), provided any of these coordinates is not outside the park.
You are given a matrix X of NxM size that tells us the contents of each cell.
If X[x1][y1] == X[x2][y2] then Tom can move from (x1,y1) to (x2,y2) without energy drinks.
If X[x1][y1] != X[x2][y2], then Tom can move from (x1,y1) to (x2,y2) by consuming a bottle of energy drink.
Find the minimum number of energy drinks that Tom must consume to reach (N-1,M-1) starting from (0,0).The input consists of two lines.
The first line consists of two integers N and M representing the order of the park. The next N lines each contain M characters representing the map of the park.
<b>Constraints</b>
2 <= N <= 1000
2 <= M <= 1000
'a' <= X[i][j] <= 'z'Find the minimum number of energy drinks that Tom must consume to reach (N-1,M-1) starting from (0,0).Input
2 2
aa
aa
Output
0
Input
2 3
abc
def
Output
3, 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 = 1e3 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
char c[N][N];
int dp[N][N];
signed main() {
IOS;
int n, m; cin >> n >> m;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
cin >> c[i][j], dp[i][j] = inf;
int x[] = {-1, 1, 0, 0};
int y[] = {0, 0, -1, 1};
deque<pi> q;
q.push_back({1, 1});
dp[1][1] = 0;
while(!q.empty()){
pi p = q.front();
q.pop_front();
for(int i = 0; i < 4; i++){
int nx = p.fi + x[i];
int ny = p.se + y[i];
if(nx <= 0 || ny <= 0 || nx > n || ny > m)
continue;
if(c[nx][ny] == c[p.fi][p.se] && dp[nx][ny] > dp[p.fi][p.se]){
dp[nx][ny] = dp[p.fi][p.se];
q.push_front({nx, ny});
}
if(c[nx][ny] != c[p.fi][p.se] && dp[nx][ny] > 1 + dp[p.fi][p.se]){
dp[nx][ny] = 1 + dp[p.fi][p.se];
q.push_back({nx, ny});
}
}
}
cout << dp[n][m];
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n people in a line queuing to buy tickets, where the 0th person is at the front of the line and the (n - 1)th person is at the back of the line.
You are given a 0-indexed integer array tickets of length n where the number of tickets that the ith person would like to buy is tickets[i].
Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.First line: n and k, denoting the total number of people and k.
Second line: n integers in which the ith integer denote the number of tickets ith person wants to buy.
<b>Constraints</b>
1 ≤ n ≤ 100
1 ≤ tickets[i] ≤ 100
0 ≤ k < n
Print the time taken for the person at position k (0-indexed) to finish buying tickets.
Input:
3 2
2 3 2
Output:
6
Explanation:
- In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].
- In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].
The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int k=sc.nextInt();
sc.nextLine();
int tickets[]=new int[n];
for(int i=0;i<n;i++){
tickets[i]=sc.nextInt();
}
Queue<int[]> queue = new LinkedList<>(); // Queue of people, each represented as an int array [position, ticketsRemaining]
// Initialize queue with people and their remaining tickets
for (int i = 0; i < n; i++) {
queue.offer(new int[]{i, tickets[i]});
}
int time = 0;
while (!queue.isEmpty()) {
int[] person = queue.poll();
int position = person[0];
int ticketsRemaining = person[1];
// Buy one ticket
ticketsRemaining--;
time++;
if (ticketsRemaining > 0) {
// Person still has tickets left, enqueue them at the end of the queue
queue.offer(new int[]{position, ticketsRemaining});
} else if (position == k) {
// Person at position k has finished buying tickets, return the total time elapsed
break;
}
}
System.out.print(time);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
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 br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
st = new StringTokenizer(br.readLine());
String s = st.nextToken();
int curzeroes = 0;
StringBuilder sb = new StringBuilder();
int len = s.length();
for(int i = 0;i<len;i++){
if(s.charAt(i) == '1'){
if(curzeroes == 0){
sb.append("1");
}
else{
curzeroes--;
}
}
else{
curzeroes++;
}
}
for(int i = 0;i<curzeroes;i++){
sb.append("0");
}
if(sb.length() == 0 && curzeroes == 0){
bw.write("-1\n");
}
else{
bw.write(sb.toString()+"\n");
}
bw.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: arr = input()
c = 0
res = ""
n =len(arr)
for i in range(n):
if arr[i]=='0':
c+=1
else:
if c==0:
res+='1'
else:
c-=1
for i in range(c):
res+='0'
if len(res)==0:
print(-1)
else:
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, 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;
string s; cin >> s;
stack<int> st;
for(int i = 0; i < (int)s.length(); i++){
if(!st.empty() && s[st.top()] == '0' && s[i] == '1'){
st.pop();
}
else
st.push(i);
}
string res = "";
while(!st.empty()){
res += s[st.top()];
st.pop();
}
reverse(res.begin(), res.end());
if(res == "") res = "-1";
cout << res;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main{
public static void main(String args[])throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(in.readLine());
int x, y, start, end, num=n, firstcond, secondcond, thirdcond, fourthcond, flag;
if(n%2!=0)
num++;
num = num/2;
int[][] a = new int[n][n];
for(x=0; x<n; x++){
String nextLine[] = in.readLine().split(" ");
for(y=0; y<n; y++){
a[x][y] = Integer.parseInt(nextLine[y]);
}
}
start = 0;
end = n-1;
while(num>=1){
flag=0;
firstcond = secondcond = thirdcond = fourthcond = 1;
for(x=start, y=start;
(firstcond==1 || secondcond==1 || thirdcond==1 || fourthcond==1) && (x>=start && y>=start && x<=end && y<=end);
){
System.out.print(a[x][y] + " ");
if(firstcond==1){
x++;
if(x==end+1){
firstcond=0;
x--;
y++;
}
}else if(secondcond==1){
y++;
if(y==end+1){
secondcond=0;
y--;
x--;
}
}else if(thirdcond==1){
x--;
if(x==start-1){
if(flag==0)
break;
thirdcond=0;
x++;
y--;
}
flag=1;
}else{
y--;
if(y==start){
fourthcond=0;
x++;
y++;
}
}
}
start++;
end--;
num--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: n = int(input())
arr = []
for i in range(n):
j = input().split()
arr.append([int(xx) for xx in j])
def counterClockspiralPrint(m, n, arr) :
k = 0; l = 0
cnt = 0
total = m * n
while (k < m and l < n) :
if (cnt == total) :
break
for i in range(k, m) :
print(arr[i][l], end = " ")
cnt += 1
l += 1
if (cnt == total) :
break
for i in range (l, n) :
print( arr[m - 1][i], end = " ")
cnt += 1
m -= 1
if (cnt == total) :
break
if (k < m) :
for i in range(m - 1, k - 1, -1) :
print(arr[i][n - 1], end = " ")
cnt += 1
n -= 1
if (cnt == total) :
break
if (l < n) :
for i in range(n - 1, l - 1, -1) :
print( arr[k][i], end = " ")
cnt += 1
k += 1
counterClockspiralPrint(n, n, arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define N 1000
void counterClockspiralPrint( int m,
int n,
int arr[][N])
{
int i, k = 0, l = 0;
// k - starting row index
// m - ending row index
// l - starting column index
// n - ending column index
// i - iterator
// initialize the count
int cnt = 0;
// total number of
// elements in matrix
int total = m * n;
while (k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
cout << arr[i][l] << " ";
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
cout << arr[m - 1][i] << " ";
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
cout << arr[i][n - 1] << " ";
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
cout << arr[k][i] << " ";
cnt++;
}
k++;
}
}
}
int main()
{
int n;
cin>>n;
int arr[n][N];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
counterClockspiralPrint(n,n, arr);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: function printAntiClockWise(mat)
{
let i, k = 0, l = 0;
let m = N;
let n = N;
let cnt = 0, total = m*n;
while(k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
process.stdout.write(mat[i][l] + " ");
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
process.stdout.write(mat[m - 1][i] + " ");
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
process.stdout.write(mat[i][n - 1] + " ");
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
process.stdout.write(mat[k][i] + " ");
cnt++;
}
k++;
}
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: function supermarket(prices, n, k) {
// write code here
// do not console.log the answer
// return sorted array
const newPrices = prices.sort((a, b) => a - b).slice(2)
let kk = k
let price = 0;
let i = 0
while (kk-- && i < newPrices.length) {
price += newPrices[i]
i += 1
}
return price
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, 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 str[]= br.readLine().trim().split(" ");
int n=Integer.parseInt(str[0]);
int k=Integer.parseInt(str[1]);
str= br.readLine().trim().split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=Integer.parseInt(str[i]);
Arrays.sort(arr);
long sum=0;
for(int i=2;i<k+2;i++)
sum+=arr[i];
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: a=input().split()
for i in range(len(a)):
a[i]=int(a[i])
b=input().split()
for i in range(len(b)):
b[i]=int(b[i])
b.sort()
b.reverse()
b.pop()
b.pop()
s=0
while a[1]>0:
s+=b.pop()
a[1]-=1
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Toros went to the supermarket to buy K items. There were a total of N items. Each item had a distinct price P<sub>i</sub>. He is cost-effective, so he would buy the K cheapest item. But he knows that the more cheaper an item is, the more is the chances that it can be defective. So he planned to ignore 2 cheapest items and buy K from the remaining ones.
Find the total cost of all items that he would buy.The first line contains two integers N and K, denoting the total number of items in the supermarket and the number of items Toros is going to buy.
The second line contains N distinct integers P<sub>i</sub> , denoting the prices of the items
<b>Constraints:</b>
1 <= N <= 100000
1 <= K <= N - 2
1 <= P<sub>i</sub> <= 1000000Print a single integer denoting the total cost of items Toros would buy.Sample Input:
5 2
4 1 2 3 5
Sample Output:
7
Explanation:
Toros will ignore items with price 1 and 2 and would buy items with price 4 and 3.
Sample Input:
10 8
99 56 50 93 47 36 65 25 87 16
Sample Output:
533, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
int ans = 0;
for(int i = 3; i <= 2 + k; i++)
ans += a[i];
cout << ans << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
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 for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(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 for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: N = int(input())
Nums = list(map(int,input().split()))
f = False
for n in Nums:
if n < 0:
f = True
break
if (f):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
bool win=false;
for(int i=0;i<n;i++){
cin>>a;
if(a<0){win=true;}}
if(win){
cout<<"Yes";
}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, check if it contains any negative integer.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
-10000 < = Arr[i] < = 10000Print "Yes" if the array contains any negative integer else print "No".Sample Input:-
4
1 2 3 -3
Sample Output:-
Yes
Sample Input:-
3
1 2 3
Sample Output:-
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]<0){System.out.print("Yes");return;}
}
System.out.print("No");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
nr = list(map(int,input().split()))
n = nr[0]
r = nr[1]
_ = list(map(int,input().split()))
num1 = _[0]
num2 = _[1]
gcd = find_gcd(num1, num2)
for i in range(2,n):
gcd = find_gcd(gcd, _[i])
if gcd>=r:
print(gcd)
else:
while r:
num = 0
idx = None
for i in range(n):
if _[i]%r != 0:
num += 1
idx = i
if num > 1:
break
if idx!= None:
temp = _[idx]
if num == 1:
_[idx] = r
num1 = _[0]
num2 = _[1]
tgcd = find_gcd(num1, num2)
for i in range(2,n):
tgcd = find_gcd(tgcd, _[i])
gcd = max(tgcd,gcd)
_[idx] = temp
break
r -= 1
print(gcd), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main {
public static long mod = 1000000007;
public static long mod2 = 998244353;
public static void main(String[] args) throws java.lang.Exception {
Reader sc = new Reader();
FastNum in = new FastNum(System.in);
Writer out = new Writer(System.out);
int n = in.nextInt();
int r = in.nextInt();
int[] a = in.nextIntArray(n);
int ans = a[0];
for (int i = 1; i < n; i++) {
ans = gcd(ans, a[i]);
}
int max = MaxGCD(a, n);
for (int i = 1; i <= r; i++) {
ans = Math.max(ans, gcd(max, i));
}
out.println(ans);
out.flush();
}
static int MaxGCD(int a[], int n) {
int[] Prefix = new int[n + 2];
int[] Suffix = new int[n + 2];
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1) {
Prefix[i] = gcd(Prefix[i - 1],
a[i - 1]);
}
Suffix[n] = a[n - 1];
for (int i = n - 1; i >= 1; i -= 1) {
Suffix[i] = gcd(Suffix[i + 1],
a[i - 1]);
}
int ans = Math.max(Suffix[2], Prefix[n - 1]);
for (int i = 2; i < n; i += 1) {
ans = Math.max(ans, gcd(Prefix[i - 1],
Suffix[i + 1]));
}
return ans;
}
static long modSum(long p, long q) {
if (q > 0) {
return (p + q) % mod;
} else {
return (p + q + mod) % mod;
}
}
static long invMod(long p, long q, long m) {
long expo = m - 2;
while (expo != 0) {
if ((expo & 1) == 1) {
p = (p * q) % m;
}
q = (q * q) % m;
expo >>= 1;
}
return p;
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long power(long a, long b) {
if (b == 0)
return 1;
long answer = power(a, b / 2) % mod;
answer = (answer * answer) % mod;
if (b % 2 != 0)
answer = (answer * a) % mod;
return answer;
}
public static void swap(int x, int y) {
int t = x;
x = y;
y = t;
}
public static long min(long a, long b) {
if (a < b) return a;
return b;
}
public static long divide(long a, long b) {
return (a % mod * (power(b, mod - 2) % mod)) % mod;
}
public static long nCr(long n, long r) {
long answer = 1;
long k = min(r, n - r);
for (long i = 0; i < k; i++) {
answer = (answer % mod * (n - i) % mod) % mod;
answer = divide(answer, i + 1);
}
return answer % mod;
}
public static boolean plaindrome(String str) {
StringBuilder sb = new StringBuilder();
sb.append(str);
return (str.equals((sb.reverse()).toString()));
}
public static class Pair {
int a;
int b;
Pair(int s, int e) {
a = s;
b = e;
}
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair pair = (Pair) o;
return a == pair.a && b == pair.b;
}
@Override
public int hashCode() {
return Objects.hash(a, b);
}
}
static class Assert {
static void check(boolean e) {
if (!e) {
throw new AssertionError();
}
}
}
static class FastNum implements AutoCloseable {
InputStream is;
byte buffer[] = new byte[1 << 16];
int size = 0;
int pos = 0;
FastNum(InputStream is) {
this.is = is;
}
int nextChar() {
if (pos >= size) {
try {
size = is.read(buffer);
} catch (IOException e) {
throw new IOError(e);
}
pos = 0;
if (size == -1) {
return -1;
}
}
Assert.check(pos < size);
int c = buffer[pos] & 0xFF;
pos++;
return c;
}
int nextInt() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
int n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Integer.MIN_VALUE / 10
|| n == Integer.MIN_VALUE / 10 && d <= -(Integer.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
int n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Integer.MAX_VALUE / 10 || n == Integer.MAX_VALUE / 10 && d <= Integer.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
char nextCh() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
return (char) c;
}
long nextLong() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
long n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Long.MIN_VALUE / 10
|| n == Long.MIN_VALUE / 10 && d <= -(Long.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
long n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Long.MAX_VALUE / 10 || n == Long.MAX_VALUE / 10 && d <= Long.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
char[] nextCharArray(int n) {
char[] arr = new char[n];
for (int i = 0; i < n; i++) {
arr[i] = nextCh();
}
return arr;
}
@Override
public void close() {
}
}
static class Writer extends PrintWriter {
public Writer(java.io.Writer out) {
super(out);
}
public Writer(java.io.Writer out, boolean autoFlush) {
super(out, autoFlush);
}
public Writer(OutputStream out) {
super(out);
}
public Writer(OutputStream out, boolean autoFlush) {
super(out, autoFlush);
}
public void printArray(int[] arr) {
for (int j : arr) {
print(j);
print(' ');
}
println();
}
public void printArray(long[] arr) {
for (long j : arr) {
print(j);
print(' ');
}
println();
}
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final 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();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 1e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
vi divisor(N + 1);
readb(n, r);
FORD (i, r, 1)
{
for (int j = i; j <= N; j += i)
if (!divisor[j]) divisor[j] = i;
}
readarr(a, n);
int pre[n + 1] = {}, suf[n + 2] = {};
FOR (i, 1, n) pre[i] = __gcd(pre[i - 1], a[i]);
FORD (i, n, 1) suf[i] = __gcd(suf[i + 1], a[i]);
int ans = pre[n];
FOR (i, 1, n)
{
int x = __gcd(pre[i - 1], suf[i + 1]);
ans = max(ans, divisor[x]);
}
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N points on 2D plane, you have to setup a camp at a point such that sum of Manhattan distance all the points from that point is minimum. If there are many such points you have to find the point with minimum X coordinate and if there are many points with same X coordinate, you have to minimize Y coordinate.
Manhattan distance between points (x1, y1) and (x2, y2) = |x1 - x2| + |y1 - y2|.First line of input contains N.
Next N lines contains two space separated integers denoting the ith coordinate.
Constraints:
1 <= N <= 100000
1 <= X[i], Y[i] <= 1000000000
Note:- the camp can overlap with the given points and the given points can also overlap(you have to consider overlapping points separately).Print two space separated integers, denoting the X and Y coordinate of the camp.Sample Input
3
3 3
1 1
3 2
Sample Output
3 2
Explanation:
Sum of distances = 1 + 3 + 0 = 4
This is the minimum distance possible., 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());
int arr[] = new int[n];
int brr[] = new int[n];
for(int i=0;i<n;i++){
String str[] = br.readLine().split(" ");
arr[i] = Integer.parseInt(str[0]);
brr[i] = Integer.parseInt(str[1]);
}
Arrays.sort(arr);
Arrays.sort(brr);
int m=0;
if(n%2==0){
m = n/2;
}else{
m = (n/2)+1;
}
System.out.println(arr[m-1]+" "+brr[m-1]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N points on 2D plane, you have to setup a camp at a point such that sum of Manhattan distance all the points from that point is minimum. If there are many such points you have to find the point with minimum X coordinate and if there are many points with same X coordinate, you have to minimize Y coordinate.
Manhattan distance between points (x1, y1) and (x2, y2) = |x1 - x2| + |y1 - y2|.First line of input contains N.
Next N lines contains two space separated integers denoting the ith coordinate.
Constraints:
1 <= N <= 100000
1 <= X[i], Y[i] <= 1000000000
Note:- the camp can overlap with the given points and the given points can also overlap(you have to consider overlapping points separately).Print two space separated integers, denoting the X and Y coordinate of the camp.Sample Input
3
3 3
1 1
3 2
Sample Output
3 2
Explanation:
Sum of distances = 1 + 3 + 0 = 4
This is the minimum distance possible., I have written this Solution Code: n = int(input())
x = []
y = []
for _ in range(n):
tempX,tempY = map(int,input().split())
x.append(tempX)
y.append(tempY)
x.sort()
y.sort()
print(x[(n-1)//2], y[(n-1)//2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N points on 2D plane, you have to setup a camp at a point such that sum of Manhattan distance all the points from that point is minimum. If there are many such points you have to find the point with minimum X coordinate and if there are many points with same X coordinate, you have to minimize Y coordinate.
Manhattan distance between points (x1, y1) and (x2, y2) = |x1 - x2| + |y1 - y2|.First line of input contains N.
Next N lines contains two space separated integers denoting the ith coordinate.
Constraints:
1 <= N <= 100000
1 <= X[i], Y[i] <= 1000000000
Note:- the camp can overlap with the given points and the given points can also overlap(you have to consider overlapping points separately).Print two space separated integers, denoting the X and Y coordinate of the camp.Sample Input
3
3 3
1 1
3 2
Sample Output
3 2
Explanation:
Sum of distances = 1 + 3 + 0 = 4
This is the minimum distance possible., 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 n;
cin>>n;
int x[n],y[n];
for(int i=0;i<n;++i){
cin>>x[i]>>y[i];
}
sort(x,x+n);
sort(y,y+n);
cout<<x[(n-1)/2]<<" "<<y[(n-1)/2];
#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: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: n = int(input())
elements = input()
array = str.split(elements)
for i in range(len(array)):
array[i] = int(array[i])
array.sort()
sum1 = 0
sum2 = 0
for x in range(int(n/2)):
sum1 += array[x]
for x in range(int(n/2)):
sum2 += array[x+int((n/2))]
diff = sum2 - sum1
print(diff), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
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_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin >> n;
assert(n%2 == 0);
vll A(n);
REP(i, 0, n)
{
cin >> A[i];
}
sort(all(A));
ll big = 0, small = 0;
REP(i, 0, n)
{
if(i<n/2) small += A[i];
else big += A[i];
}
cout << big - small << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt();
long arr[] = sc.readArrayLong(n);
ruffleSort(arr);
long a = 0;
for(int i = 0; i<n/2; i++) {
a+=arr[i];
}
long b = 0;
for(int i = n/2; i<n; i++) {
b+=arr[i];
}
System.out.println(abs(b-a));
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers. Find the number of pairs (i, j) of integers satisfying all of the following conditions:
1 ≤ i < j ≤ N
A<sub>i</sub> ≠ A<sub>j</sub>
The first line of the input consists of an integer N. The second line consists of N space separated integers.
<b>Constraints</b>
All values in input are integers.
2 ≤ N ≤ 3×10^5
1 ≤ A<sub>i</sub> ≤ 10^9Print the answer as an integer.<b>Sample Input 1</b>
3
1 7 1
<b>Sample Output 1</b>
2
<b>Sample Input 2</b>
10
1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000
<b>Sample Output 2</b>
45, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
vector<int> a(n);
for(auto &nx : a){cin >> nx;}
long long res1=(n*(n-1)/2),res2=0;
map<int,long long> mp;
for(int i=0;i<n;i++){mp[a[i]]++;}
for(int i=0;i<n;i++){res2+=(n-mp[a[i]]);}
res2/=2;
sort(a.begin(),a.end());
a.push_back(-1);
long long cnt=1;
for(int i=0;i<n;i++){
if(a[i]!=a[i+1]){
res1-=((cnt*(cnt-1))/2);
cnt=1;
}
else{cnt++;}
}
assert(res1==res2);
cout << res1 << '\n';
//cout << res2 << '\n';
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Print the integer just less than equal to N which has atleast one among {5 , 7, 11} as a divisor.The input contain only the integer N.
Constraints
5 <= N <= 1000000000Print an integer just less than equal to N, which is a multiple of 5 ,7 or 11.Sample Input
13
Sample Output
11
Sample Input
16
Sample Output
15
Explanation:-
For testcase1:
13 is not divisible by 5 or 7 or 11
12 is not divisible by 5 or 7 or 11
11 is divisible by 11 so ans is 11, 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));
long num = Long.parseLong(br.readLine().trim());
System.out.print(checkNum(num));
}
private static long checkNum(long num){
if(num%5==0 || num%7==0 || num%11==0){
return num;
}
return checkNum(num-1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Print the integer just less than equal to N which has atleast one among {5 , 7, 11} as a divisor.The input contain only the integer N.
Constraints
5 <= N <= 1000000000Print an integer just less than equal to N, which is a multiple of 5 ,7 or 11.Sample Input
13
Sample Output
11
Sample Input
16
Sample Output
15
Explanation:-
For testcase1:
13 is not divisible by 5 or 7 or 11
12 is not divisible by 5 or 7 or 11
11 is divisible by 11 so ans is 11, I have written this Solution Code: n = int(input())
x = 5
y = 7
z = 11
print(max((n//x)*x,max((n//y)*y,(n//z)*z))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Print the integer just less than equal to N which has atleast one among {5 , 7, 11} as a divisor.The input contain only the integer N.
Constraints
5 <= N <= 1000000000Print an integer just less than equal to N, which is a multiple of 5 ,7 or 11.Sample Input
13
Sample Output
11
Sample Input
16
Sample Output
15
Explanation:-
For testcase1:
13 is not divisible by 5 or 7 or 11
12 is not divisible by 5 or 7 or 11
11 is divisible by 11 so ans is 11, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main()
{
long n;
cin>>n;
long x=5,y=7,z=11;
cout<<max((n/x)*x,max((n/y)*y,(n/z)*z));
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%n==0):
print("Yes")
else:
print("No");, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, 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 n,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#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 an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
Long m = sc.nextLong();
if(n%m==0){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: def NthNumber(N):
return 24+(N-1)*13
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<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>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: static int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args){
FastReader read = new FastReader();
int a = read.nextInt();
int b = read.nextInt();
System.out.println(a+b);
System.out.println(a-b);
System.out.println(a*b);
System.out.println(a/b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
InputStreamReader inr = new InputStreamReader(System.in);
br = new BufferedReader(inr);
}
String next(){
while(st==null || !st.hasMoreElements())
try{
st = new StringTokenizer(br.readLine());
}
catch(IOException e){
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
long nextLong(){
return Long.parseLong(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch(IOException e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: a,b=map(int,input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice and Bob are playing a game where Bob gives Alice an array <b>A</b> of size N. There are some positions in the array where the index (According to 1-based indexing) and the element at that index both are prime numbers. Alice has to find those elements and give that to Bob in order of their appearance in the array. Your task is to help Alice complete his task and print the elements.The first line of the input contains an integer N representing the number of elements in the array that Bob gave.
The second line of the input contains n space-separated elements of the array <b>A</b>.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>4</sup>
1 ≤ A[i] ≤ 10<sup>4</sup>Print the required elements separated by space in a single line.<b>Sample Input</b>
10
2 3 5 7 11 13 17 19 23 29
<b>Sample Output</b>
3 5 11 17
<b>Explanation</b>
The array indexing is [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] so the prime number according to the array indexing are 2, 3, 5, 7, and elements belonging to respective indexes are 3, 5, 11, 17. Hence the elements which are prime and are also at prime indexes are 3, 5, 11, and 17., I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
// operator overload of << for vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
for (const auto &x : v)
os << x << " ";
return os;
}
bool prime[10001];
void SieveOfEratosthenes(int n) {
memset(prime, true, sizeof(prime));
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
void BobsPrimeArray(int n, vector<int> &arr) {
SieveOfEratosthenes(10000);
bool ansFound = false;
for (int i = 1; i < n; ++i) {
if (arr[i] != 1 && arr[i]!=0 && prime[i + 1] && prime[arr[i]]) {
cout << arr[i] << " ";
ansFound = true;
}
}
if (!ansFound) {
cout << "-1\n";
}
}
int32_t main() {
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; ++i) {
cin >> arr[i];
}
BobsPrimeArray(n, arr);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: a,b,c=[int(a) for a in input().split()]
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int p = scanner.nextInt();
int tm = scanner.nextInt();
int r = scanner.nextInt();
int intrst = MaxInteger(p,tm,r);
System.out.println(intrst);
}
static int MaxInteger(int a ,int b, int c){
if(a>=b && a>=c){return a;}
if(b>=a && b>=c){return b;}
return c;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given any number S ending with 1. What will be the last 2 digits of the resultant number when raised to the power P?The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consists of string S and power P.
Constraints:
1 <= T <= 2*10^3
1 <= |S| <= 60
1 <= P <= 10^6Print the last 2 digit of the resultant. If answer is 1 digit number X, print 0X.Input:
2
1231 213
43591 194
Output:
91
61
Explanation:
Testcase 1: When (1231)^213 is calculated the last 2 digits appeared to be 91., I have written this Solution Code: t = int(input())
while (t):
n, e = map(int, input().split())
d = (((n // 10) % 10) * (e % 10)) % 10
print(d, 1, sep='')
t -= 1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given any number S ending with 1. What will be the last 2 digits of the resultant number when raised to the power P?The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consists of string S and power P.
Constraints:
1 <= T <= 2*10^3
1 <= |S| <= 60
1 <= P <= 10^6Print the last 2 digit of the resultant. If answer is 1 digit number X, print 0X.Input:
2
1231 213
43591 194
Output:
91
61
Explanation:
Testcase 1: When (1231)^213 is calculated the last 2 digits appeared to be 91., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#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 sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
string s;
cin>>s;
int yy;
cin>>yy;
int n=s.size();
if(n==1) {
cout<<"01"<<endl;
continue;
}
int p=s[n-2]-'0';
int q=(p*yy)%10LL;
//if(q==0 ) cout<<1<<endl;
cout<<q<<1<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Two numbers are represented in Linked List such that each digit corresponds to a node in the linked list. Your task is to add these two numbers and return the sum in a linked list.
<b>Note:-</b> Linked list representation of a number is from left to right i.e if the number is 123 then in the linked list it is represented as 3->2->1<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>addNumber()</b> that takes head nodes of both the linked lists as parameters.
<b>Constraints:</b>
1 <=numbers<=10<sup>1000</sup>Return the head of modified linked list.Sample Input:-
1234 45643
Sample Output:-
46877, I have written this Solution Code: static Node addNumber(Node first, Node second)
{ Node res = null; // res is head node of the resultant list
Node prev = null;
Node temp = null;
int carry = 0, sum;
while (first != null || second != null) //while both lists exist
{
sum = carry + (first != null ? first.data : 0)
+ (second != null ? second.data : 0);
carry = (sum >= 10) ? 1 : 0;
sum = sum % 10;
temp = new Node(sum);
if (res == null) {
res = temp;
} else // If this is not the first node then connect it to the rest.
{
prev.next = temp;
}
prev = temp;
if (first != null) {
first = first.next;
}
if (second != null) {
second = second.next;
}
}
if (carry > 0) {
temp.next = new Node(carry);
}
// return head of the resultant list
return res;
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., I have written this Solution Code: def FindIt(n):
sqrt_n = int(K**0.5)
check = -1
for i in range(sqrt_n - 1, sqrt_n - 2, -1):
check = i**2 + (i * 3)
if check == n:
return i
return -1
K = int(input())
print(FindIt(K)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve(){
int n;
cin >> n;
int l = 1, r = 1e9, ans = -1;
while(l <= r){
int m = (l + r)/2;
int val = m*m + 3*m;
if(val == n){
ans = m;
break;
}
if(val < n){
l = m + 1;
}
else r = m - 1;
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., 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));
long K = Long.parseLong(br.readLine());
long ans = -1;
for(long x =0;((x*x)+(3*x))<=K;x++){
if(K==((x*x)+(3*x))){
ans = x;
break;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code: n = int(input())
arr = []
x = 4
prod = 1
a = input().split()
for i in range(0,n):
prod = prod*int(a[i])
if prod % 4 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int two=0,four=0;
for(int i=0;i<n;i++){
if(a[i]%2==0){
two++;
}
if(a[i]%4==0){
four++;
}
}
if(four>=1 || two>=2){
System.out.print("YES");
}
else{
System.out.print("NO");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 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(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 int long long
#define ld 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
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// 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
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
int ct = 0;
For(i, 0, n){
int a; cin>>a;
while(a%2==0){
ct++;
a/=2;
}
}
if(ct>=2){
cout<<"YES";
}
else{
cout<<"NO";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes and an integer K, your task is to delete the Kth node from the end of the linked list<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>deleteElement()</b> that takes head node and K as parameter.
Constraints:
1 <=K<=N<= 1000
1 <=Node.data<= 1000Return the head of the modified linked listInput 1:
5 3
1 2 3 4 5
Output 1:
1 2 4 5
Explanation:
After deleting 3rd node from the end of the linked list, 3 will be deleted and the list will be as 1, 2, 4, 5.
Input 2:-
5 5
8 1 8 3 6
Output 2:-
1 8 3 6 , I have written this Solution Code: public static Node deleteElement(Node head,int k) {
int cnt=0;
Node temp=head;
while(temp!=null){
cnt++;
temp=temp.next;
}
temp=head;
k=cnt-k;
if(k==0){head=head.next; return head;}
temp=head;
cnt=0;
while(cnt!=k-1){
temp=temp.next;
cnt++;
}
temp.next=temp.next.next;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, 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 n;
while(t>0) {
n = Integer.parseInt(br.readLine());
int a[] = new int[n];
String s = br.readLine();
String arr[] = s.split(" ");
for(int i=0;i<n;i++) {
a[i] = Integer.parseInt(arr[i]);
}
int b[] = sort(a);
for(int i=0;i<n;i++) {
System.out.print(b[i] + " ");
}
System.out.println();
t--;
}
}
static int[] sort(int[] a) {
int temp, index;
for(int i=0;i<a.length-1;i++) {
index = i;
for(int j=i+1;j<a.length;j++) {
if(a[j]<a[index]) {
index = j;
}
}
temp = a[i];
a[i] = a[index];
a[index] = temp;
}
return a;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, I have written this Solution Code: for _ in range(int(input())):
n = input()
res = map(str, sorted(list( map(int,input().split()))))
print(' '.join(res)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, 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;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the heads of two singly linked lists head1 and head2, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null
For example, the following two linked lists begin to intersect at node c1<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>intersection()</b> that takes the head node of both lists as a parameter.
The first line of the test case contains:
<ul>
<li>the size of the 1st linked list</li>
<li>size of 2nd linked list</li>
<li>the size of the common part of two linked lists</li>
</ul>
The rest of the test case contains:
<ul>
<li>elements of 1st linked list</li>
<li>elements of 2nd linked list</li>
<li>common part of two linked lists</li>
</ul>Return the node of intersectionSample Input:-
4 3 4
1 2 3 4
5 6 7
9 10 11 12
Sample Output:-
9
Explanation:
1 -> 2 -> 3 -> 4
      |
       9 -> 10 -> 11 -> 12
      |
  5 -> 6 -> 7, I have written this Solution Code:
public static Node intersection(Node head1,Node head2){ int aLength = getLength(head1), bLength = getLength(head2);
Node currA = head1, currB = head2;
if (bLength > aLength)
for (int i = 0; i < bLength - aLength; i++) currB = currB.next;
if (aLength > bLength)
for (int i = 0; i < aLength - bLength; i++) currA = currA.next;
while (currA != null && currB != null) {
if (currA == currB) return currA;
currA = currA.next;
currB = currB.next;
}
return null;
}
private static int getLength(Node head) {
Node curr = head;
int len = 0;
while (curr != null) {
curr = curr.next;
len++;
}
return len;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args){
FastReader read = new FastReader();
int a = read.nextInt();
int b = read.nextInt();
System.out.println(a+b);
System.out.println(a-b);
System.out.println(a*b);
System.out.println(a/b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
InputStreamReader inr = new InputStreamReader(System.in);
br = new BufferedReader(inr);
}
String next(){
while(st==null || !st.hasMoreElements())
try{
st = new StringTokenizer(br.readLine());
}
catch(IOException e){
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
long nextLong(){
return Long.parseLong(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch(IOException e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: a,b=map(int,input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, 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().toString();
int n=Integer.parseInt(s);
int ch=0;
if(n>0){
ch=1;
}
else if(n<0) ch=-1;
switch(ch){
case 1: System.out.println("Positive");break;
case 0: System.out.println("Zero");break;
case -1: System.out.println("Negative");break;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: n = input()
if '-' in list(n):
print('Negative')
elif int(n) == 0 :
print('Zero')
else:
print('Positive'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch (num > 0)
{
// Num is positive
case 1:
printf("Positive");
break;
// Num is either negative or zero
case 0:
switch (num < 0)
{
case 1:
printf("Negative");
break;
case 0:
printf("Zero");
break;
}
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<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>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton wants to take revenge from two apples fallen on his head. So, he applies force F<sub>1</sub> on first apple (mass M<sub>1</sub>) resulting in acceleration of A<sub>1</sub> and F<sub>2</sub> on second apple (mass M<sub>2</sub>) resulting in acceleration of A<sub>2</sub>. Given M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>. Calculate total force applied by him on two apples.
<b>Note:</b> F = M*A is the equation of relation between force, mass and acceleration.First line contains four integers M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>.
1 <= M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub> <= 100Output total force applied by Newton.INPUT:
1 2 3 4
OUTPUT:
14
Explanation:
Total force is equal to 1*2 + 3*4 = 14., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int m1,a1,m2,a2;
cin >> m1 >> a1 >> m2 >> a2;
cout << (m1*a1)+(m2*a2) << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Cf cf = new Cf();
cf.solve();
}
static class Cf {
InputReader in = new InputReader(System.in);
OutputWriter out = new OutputWriter(System.out);
int mod = (int)1e9+7;
public void solve() {
int t = in.readInt();
if(t>=6) {
out.printLine("No");
}else {
out.printLine("Yes");
}
}
public long findPower(long x,long n) {
long ans = 1;
long nn = n;
while(nn>0) {
if(nn%2==1) {
ans = (ans*x) % mod;
nn-=1;
}else {
x = (x*x)%mod;
nn/=2;
}
}
return ans%mod;
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
private static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
private 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]);
}
writer.flush();
}
public void printLine(Object... objects) {
print(objects);
writer.println();
writer.flush();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, 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();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
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 mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n<6)
cout<<"Yes";
else
cout<<"No";
#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: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: n=int(input())
if(n>=6):
print("No")
else:
print("Yes"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a list Arr[] of integers representing a list compressed with run-length encoding.
Consider each adjacent pair of elements [freq, val] = [Arr[2*i], Arr[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a subarray. Concatenate all the subarray from left to right to generate the decompressed array.
Print the decompressed array.The input line contains T, denoting the number of testcases. Each testcase contains 2 lines. The first line contains size of array. Second line contains elements of array separated by space.
Note: size of input array is even
Constraints:
1 <= T <= 100
2 <= N <= 100
0 <= Arr[i] <= 100For each testcase you need to print the decompressed array in a new line.Sample Input:
2
4
1 2 3 4
4
1 1 2 3
Sample Output:
2 4 4 4
1 3 3
Explanation:
Testcase 1: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4]., I have written this Solution Code: n = int(input())
for _ in range(n):
nt = int(input())
l = [int(x) for x in input().split()];
for i in range(0,nt,2):
print((str(l[i+1]) + " ") * l[i],end="")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a list Arr[] of integers representing a list compressed with run-length encoding.
Consider each adjacent pair of elements [freq, val] = [Arr[2*i], Arr[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a subarray. Concatenate all the subarray from left to right to generate the decompressed array.
Print the decompressed array.The input line contains T, denoting the number of testcases. Each testcase contains 2 lines. The first line contains size of array. Second line contains elements of array separated by space.
Note: size of input array is even
Constraints:
1 <= T <= 100
2 <= N <= 100
0 <= Arr[i] <= 100For each testcase you need to print the decompressed array in a new line.Sample Input:
2
4
1 2 3 4
4
1 1 2 3
Sample Output:
2 4 4 4
1 3 3
Explanation:
Testcase 1: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4]., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main(String args[])throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
int N = Integer.parseInt(read.readLine());
String str[] = read.readLine().trim().split(" ");
int arr[] = new int[N];
for(int i = 0; i < N; i++)
arr[i] = Integer.parseInt(str[i]);
int res[] = new int[decompress_EncodedArray(arr, N).length];
res = decompress_EncodedArray(arr, N);
//Collections.addAll(list, );
print(res);
System.out.println();
}
}
static void print(int list[])
{
for(int i = 0; i < list.length; i++)
System.out.print(list[i] + " ");
}
public static int [] decompress_EncodedArray(int[] nums, int n) {
int[] intArr=null;
int opArrSize=0;
int opArrIndex=0;
for(int i=0; i<nums.length; i += 2)
{
opArrSize += nums[i];
}
intArr= new int[opArrSize];
for(int j=0; j<nums.length; j += 2)
{
int freq= nums[j];
int val = nums[j+1];
while(freq>0)
{
intArr[opArrIndex] = val;
freq--;
opArrIndex++;
}
}
return intArr;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a list Arr[] of integers representing a list compressed with run-length encoding.
Consider each adjacent pair of elements [freq, val] = [Arr[2*i], Arr[2*i+1]] (with i >= 0). For each such pair, there are freq elements with value val concatenated in a subarray. Concatenate all the subarray from left to right to generate the decompressed array.
Print the decompressed array.The input line contains T, denoting the number of testcases. Each testcase contains 2 lines. The first line contains size of array. Second line contains elements of array separated by space.
Note: size of input array is even
Constraints:
1 <= T <= 100
2 <= N <= 100
0 <= Arr[i] <= 100For each testcase you need to print the decompressed array in a new line.Sample Input:
2
4
1 2 3 4
4
1 1 2 3
Sample Output:
2 4 4 4
1 3 3
Explanation:
Testcase 1: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4]., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i+=2){
while(a[i]>0){
cout<<a[i+1]<<" ";
a[i]--;
}
}
cout<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves pattern, so this time she wishes to draw a pattern as:-
*****
****
***
**
*
Since Sara does not know how to code, help her to draw this pattern.<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>Triangle()</b> that takes no parameters.Print the pattern as shown in the example.Sample Output:-
*****
****
***
**
*, I have written this Solution Code: def Triangle():
for i in range(5):
for j in range(5-i):
print("*", end='')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves pattern, so this time she wishes to draw a pattern as:-
*****
****
***
**
*
Since Sara does not know how to code, help her to draw this pattern.<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>Triangle()</b> that takes no parameters.Print the pattern as shown in the example.Sample Output:-
*****
****
***
**
*, I have written this Solution Code: static void Triangle(){
System.out.println("*****");
System.out.println("****");
System.out.println("***");
System.out.println("**");
System.out.println("*");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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