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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: Daemon has 2 dragon eggs, one is made of A grams of Gold and the other is made of B grams of Silver.
We know that Gold costs G coins per gram and Silver costs S coins per gram. You need to determine which of the two dragon eggs are more valuable.The input contains 4 space-separated integers G, S, A, B.
<b> Constraints: </b>
1 ≤ G, S, A, B ≤ 10<sup>4</sup>Print "Gold" (without quotes) if the Gold egg costs <b>equal to or more than</b> the silver egg. Otherwise, print "Silver" (without quotes).
Note that the output is case-sensitive.Sample Input 1:
5 4 4 5
Sample Output 1:
Gold
Sample Input 2:
1 1 2 3
Sample Output 2:
Silver, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader bi = new BufferedReader(new InputStreamReader(System.in));
int [] num = new int[4];
String[] str = bi.readLine().split(" ");
for(int i = 0; i < str.length; i++)
num[i] = Integer.parseInt(str[i]);
if((num[0] * num[2]) >= (num[1] * num[3]))
System.out.print("Gold");
else
System.out.print("Silver");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Daemon has 2 dragon eggs, one is made of A grams of Gold and the other is made of B grams of Silver.
We know that Gold costs G coins per gram and Silver costs S coins per gram. You need to determine which of the two dragon eggs are more valuable.The input contains 4 space-separated integers G, S, A, B.
<b> Constraints: </b>
1 ≤ G, S, A, B ≤ 10<sup>4</sup>Print "Gold" (without quotes) if the Gold egg costs <b>equal to or more than</b> the silver egg. Otherwise, print "Silver" (without quotes).
Note that the output is case-sensitive.Sample Input 1:
5 4 4 5
Sample Output 1:
Gold
Sample Input 2:
1 1 2 3
Sample Output 2:
Silver, I have written this Solution Code:
G, S, A, B = input().split()
g = int(G)
s = int(S)
a = int(A)
b = int(B)
gold = (g*a)
silver = (s*b)
if(gold>=silver):
print("Gold")
else:
print("Silver"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Daemon has 2 dragon eggs, one is made of A grams of Gold and the other is made of B grams of Silver.
We know that Gold costs G coins per gram and Silver costs S coins per gram. You need to determine which of the two dragon eggs are more valuable.The input contains 4 space-separated integers G, S, A, B.
<b> Constraints: </b>
1 ≤ G, S, A, B ≤ 10<sup>4</sup>Print "Gold" (without quotes) if the Gold egg costs <b>equal to or more than</b> the silver egg. Otherwise, print "Silver" (without quotes).
Note that the output is case-sensitive.Sample Input 1:
5 4 4 5
Sample Output 1:
Gold
Sample Input 2:
1 1 2 3
Sample Output 2:
Silver, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main() {
int g, s, a, b;
cin >> g >> s >> a >> b;
cout << ((g * a >= s * b) ? "Gold" : "Silver");
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: static void pattern(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j + " ");
}
System.out.println();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: function pattern(n) {
// write code herenum
for(let i = 1;i<=n;i++){
let str = ''
for(let k = 1; k <= i;k++){
if(k === 1) {
str += `${k}`
}else{
str += ` ${k}`
}
}
console.log(str)
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: def patternPrinting(n):
for i in range(1,n+1):
for j in range (1,i+1):
print(j,end=' ')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: function PokemonMaster(a,b) {
// write code here
// do no console.log the answer
// return the output using return keyword
const ans = (a - b >= 0) ? 1 : 0
return ans
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: def PokemonMaster(A,B):
if(A>=B):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: static int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers N and K which are part of the equation X<sub>1</sub> + X<sub>2</sub> + X<sub>3</sub> +. . + X<sub>K</sub> = N. Find the number of non-negative integral solutions of (X<sub>1</sub>, X<sub>2</sub>,. ., X<sub>K</sub>).
Since the answer can be pretty large, print answer modulo (10^9 + 7)
You may like to read about modular multiplicative inverse.Single line of input containing two space-separated integers N and K
1 <= N, K <= 100000Print the number of solutions modulo (10^9 + 7)Sample Input:
3 2
Sample Output:
4
Explanation:
The solutions are:
(0, 3)
(1, 2)
(2, 1)
(3, 0), 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 power(int a, int b){
int ans = 1;
while(b){
if(b&1)
ans = (ans*a) % mod;
b >>= 1;
a = (a*a) % mod;
}
return ans;
}
void solve(){
int x, y;
cin >> x >> y;
x = x + y - 1;
y = y - 1;
int ans = 1;
for(int i = 1; i <= y; i++){
ans = (ans*(x+1-i)) % mod;
ans = (ans*power(i, mod-2)) % mod;
}
cout << ans;
}
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 three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N containing 0s and 1s only. The task is to count the subarrays having an equal number of 0s and 1s.The first line of the input contains an integer N denoting the size of the array and the second line contains N space-separated 0s and 1s.
Constraints:-
1 <= N <= 10^6
0 <= A[i] <= 1For each test case, print the count of required sub-arrays in new line.Sample Input
7
1 0 0 1 0 1 1
Sample Output
8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6), I have written this Solution Code: size = int(input())
givenList = list(map(int,input().split()))
hs = {}
sm = 0
ct = 0
for i in givenList:
if i == 0:
i = -1
sm = sm + i
if sm == 0:
ct += 1
if sm not in hs.keys():
hs[sm] = 1
else:
freq = hs[sm]
ct = ct +freq
hs[sm] = freq + 1
print(ct), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N containing 0s and 1s only. The task is to count the subarrays having an equal number of 0s and 1s.The first line of the input contains an integer N denoting the size of the array and the second line contains N space-separated 0s and 1s.
Constraints:-
1 <= N <= 10^6
0 <= A[i] <= 1For each test case, print the count of required sub-arrays in new line.Sample Input
7
1 0 0 1 0 1 1
Sample Output
8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6), I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 1000001
int a[max1];
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]==0){a[i]=-1;}
}
long sum=0;
unordered_map<long,int> m;
long cnt=0;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==0){cnt++;}
cnt+=m[sum];
m[sum]++;
}
cout<<cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N containing 0s and 1s only. The task is to count the subarrays having an equal number of 0s and 1s.The first line of the input contains an integer N denoting the size of the array and the second line contains N space-separated 0s and 1s.
Constraints:-
1 <= N <= 10^6
0 <= A[i] <= 1For each test case, print the count of required sub-arrays in new line.Sample Input
7
1 0 0 1 0 1 1
Sample Output
8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6), I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int arrSize = sc.nextInt();
long arr[] = new long[arrSize];
for(int i = 0; i < arrSize; i++)
arr[i] = sc.nextInt();
System.out.println(countSubarrays(arr, arrSize));
}
static long countSubarrays(long arr[], int arrSize)
{
for(int i = 0; i < arrSize; i++)
{
if(arr[i] == 0)
arr[i] = -1;
}
long ans = 0;
long sum = 0;
HashMap<Long, Integer> hash = new HashMap<>();
for(int i = 0; i < arrSize; i++)
{
sum += arr[i];
if(sum == 0)
ans++;
if(hash.containsKey(sum) == true)
{
ans += hash.get(sum);
int freq = hash.get(sum);
hash.put(sum, freq+1);
}
else hash.put(sum, 1);
}
return 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: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public 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 reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, 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 mod2 = 998244353;
const ll N = 2e5 + 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()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays X and Y of size N each. These represent the coordinates of the points where the i<sup>th</sup> point is given by (X[i], Y[i]). You need to find K points that are closest to origin. The distance between two points is given by the Manhattan distance formula i. e. D = |x1- x2|+|y1- y2|
If there are more than two points with same distance and you need to choose among them, choose the one which appears earlier in the array.
Print all the K points.
Note: All the coordinates are integers.First line contains two integers N and K.
Next line contains N space separated integers denoting elements of array X.
Next line contains N space separated integers denoting elements of array Y.
Constraints
1 <= K <= N <= 10^5
1 <= Xi<= 10^5
1 <= Yi<= 10^5Print K lines where each line contains two space separated integers denoting the x and y coordinate of points arranged in the order of their distance in decreasing order. If two points have same distance, print the point which has higher index before printing point with same distance of lower index.Sample Input 1:
3 1
1 2 3
5 6 7
Output
1 5
Explanation:
Distance of Point 1 : (1+5) = 6
Distance of Point 2 : (2+6) = 8
Distance of Point 3 : (3+7) = 10
Print the first point as it is closest to origin.
Sample Input 2:
5 3
3 6 6 9 1
1 1 1 1 1
Output
6 1
3 1
1 1
Explanation
Distance of Point 1 : (3+1) = 4
Distance of Point 2 : (6+1) = 7
Distance of Point 3 : (6+1) = 7
Distance of Point 4 : (9+1) = 10
Distance of Point 5 : (1+1) = 2
Print the points 2, 1 and 5 as they are closest to origin.
Point 2 and 3 have same distance but we choose the point which occur earlier., 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[] line1 = br.readLine().split(" ");
String[] line2 = br.readLine().split(" ");
String[] line3 = br.readLine().split(" ");
int n = Integer.parseInt(line1[0]);
int k = Integer.parseInt(line1[1]);
int[] X = new int[line2.length];
int[] Y = new int[line3.length];
PriorityQueue<D> pq = new PriorityQueue<>(Collections.reverseOrder());
for(int a=0;a<line2.length;a++){
X[a] = Integer.parseInt(line2[a]);
Y[a] = Integer.parseInt(line3[a]);
}
for(int b=0;b<line2.length;b++){
D temp = new D(X[b]+Y[b],b);
pq.add(temp);
if(pq.size()>k){
pq.remove();
}
}
for(int b=0;b<k;b++){
D temp2 = pq.remove();
System.out.println(X[temp2.idx]+" "+Y[temp2.idx]);
}
}
}
class D implements Comparable<D>{
int dis;
int idx;
public D(int dist, int indx){
dis = dist;
idx = indx;
}
public int compareTo(D d){
if(this.dis > d.dis){
return 1;
}else if(this.dis < d.dis){
return -1;
}else{
if(this.idx > d.idx){
return 1;
}else{
return -1;
}
}
}
public String toString(){
return this.dis+" "+this.idx+" | ";
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays X and Y of size N each. These represent the coordinates of the points where the i<sup>th</sup> point is given by (X[i], Y[i]). You need to find K points that are closest to origin. The distance between two points is given by the Manhattan distance formula i. e. D = |x1- x2|+|y1- y2|
If there are more than two points with same distance and you need to choose among them, choose the one which appears earlier in the array.
Print all the K points.
Note: All the coordinates are integers.First line contains two integers N and K.
Next line contains N space separated integers denoting elements of array X.
Next line contains N space separated integers denoting elements of array Y.
Constraints
1 <= K <= N <= 10^5
1 <= Xi<= 10^5
1 <= Yi<= 10^5Print K lines where each line contains two space separated integers denoting the x and y coordinate of points arranged in the order of their distance in decreasing order. If two points have same distance, print the point which has higher index before printing point with same distance of lower index.Sample Input 1:
3 1
1 2 3
5 6 7
Output
1 5
Explanation:
Distance of Point 1 : (1+5) = 6
Distance of Point 2 : (2+6) = 8
Distance of Point 3 : (3+7) = 10
Print the first point as it is closest to origin.
Sample Input 2:
5 3
3 6 6 9 1
1 1 1 1 1
Output
6 1
3 1
1 1
Explanation
Distance of Point 1 : (3+1) = 4
Distance of Point 2 : (6+1) = 7
Distance of Point 3 : (6+1) = 7
Distance of Point 4 : (9+1) = 10
Distance of Point 5 : (1+1) = 2
Print the points 2, 1 and 5 as they are closest to origin.
Point 2 and 3 have same distance but we choose the point which occur earlier., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long int
#define f(n) for(int i=0;i<n;i++)
#define fo(n) for(int j=0;j<n;j++)
#define foo(n) for(int i=1;i<=n;i++)
#define ff first
#define ss second
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define vp vector<pii>
#define test int tt; cin>>tt; while(tt--)
#define mod 1000000007
void fastio()
{
ios_base::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("Input 1.txt", "r", stdin);
freopen("Output 1.txt", "w", stdout);
#endif
}
int main() {
fastio();
int n, k;
cin >> n >> k;
priority_queue <pair<int, int>>q;
int x[n];
int y[n];
f(n) {
cin >> x[i];
}
f(n) {
cin >> y[i];
}
f(n) {
int d = x[i] + y[i];
q.push({d, i});
if (q.size() > k) {
q.pop();
}
}
while (q.size()) {
int i = q.top().ss;
q.pop();
cout << x[i] << " " << y[i] << endl;
}
}, 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 a doubly linked list consisting of N nodes and two integers <b>P</b> and <b>K</b>. Your task is to add an element K at the Pth position from the start 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>insertnew()</b>. The description of parameters are mentioned below:
<b>head</b>: head node of the double linked list
<b>K</b>: the element which you have to insert
<b>P</b>: the position at which you have insert
Constraints:
1 <= P <=N <= 1000
1 <=K, Node.data<= 1000
In the sample Input N, P and K are in the order as mentioned below:
<b>N P K</b>Return the head of the modified linked list.Sample Input:-
5 3 2
1 3 2 4 5
Sample Output:-
1 3 2 2 4 5, I have written this Solution Code: public static Node insertnew(Node head, int k,int pos) {
int cnt=1;
if(pos==1){Node temp=new Node(k);
temp.next=head;
head.prev=temp;
return temp;}
Node temp=head;
while(cnt!=pos-1){
temp=temp.next;
cnt++;
}
Node x= new Node(k);
x.next=temp.next;
temp.next.prev=x;
temp.next=x;
x.prev=temp;
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long count(int[] arr, int l, int h, int[] aux) {
if (l >= h) return 0;
int mid = (l +h) / 2;
long count = 0;
count += count(aux, l, mid, arr);
count += count(aux, mid + 1, h, arr);
count += merge(arr, l, mid, h, aux);
return count;
}
static long merge(int[] arr, int l, int mid, int h, int[] aux) {
long count = 0;
int i = l, j = mid + 1, k = l;
while (i <= mid || j <= h) {
if (i > mid) {
arr[k++] = aux[j++];
} else if (j > h) {
arr[k++] = aux[i++];
} else if (aux[i] <= aux[j]) {
arr[k++] = aux[i++];
} else {
arr[k++] = aux[j++];
count += mid + 1 - i;
}
}
return count;
}
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
String str[];
str = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
str = br.readLine().split(" ");
int arr[] =new int[n];
for (int j = 0; j < n; j++) {
arr[j] = Integer.parseInt(str[j]);
}
int[] aux = arr.clone();
System.out.print(count(arr, 0, n - 1, aux));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: count=0
def implementMergeSort(arr,s,e):
global count
if e-s==1:
return
mid=(s+e)//2
implementMergeSort(arr,s,mid)
implementMergeSort(arr,mid,e)
count+=merge_sort_place(arr,s,mid,e)
return count
def merge_sort_place(arr,s,mid,e):
arr3=[]
i=s
j=mid
count=0
while i<mid and j<e:
if arr[i]>arr[j]:
arr3.append(arr[j])
j+=1
count+=(mid-i)
else:
arr3.append(arr[i])
i+=1
while (i<mid):
arr3.append(arr[i])
i+=1
while (j<e):
arr3.append(arr[j])
j+=1
for x in range(len(arr3)):
arr[s+x]=arr3[x]
return count
n=int(input())
arr=list(map(int,input().split()[:n]))
c=implementMergeSort(arr,0,len(arr))
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long _mergeSort(long long arr[], int temp[], int left, int right);
long long merge(long long arr[], int temp[], int left, int mid, int right);
/* This function sorts the input array and returns the
number of inversions in the array */
long long mergeSort(long long arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
long long _mergeSort(long long arr[], int temp[], int left, int right)
{
long long mid, inv_count = 0;
if (right > left) {
/* Divide the array into two parts and
call _mergeSortAndCountInv()
for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
long long merge(long long arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
long long inv_count = 0;
i = left; /* i is index for left subarray*/
j = mid; /* j is index for right subarray*/
k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;}
int main(){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];}
long long ans = mergeSort(a, n);
cout << ans; }
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He want to crush all of them. He feels that it would be boring to crush the candies randomly, so he devices a method to crush them. He divides these candies in minimum number of groups such than no group contains more than 3 candy. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much total cost would be incurred. Can you help him?1 <= N <= 10^9return the cost from the functionSample Input 1:
4
Explanation:
Query 1: First step John divides the candies in two groups of 3 and 1 candy respectively. Crushing one- one candy from both group would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: // ignore number of testcases/queries
// n is the number as provided in input
function findCost(n) {
// write code here
// do not console.log the answer
// return answer as a number
if (n == 0) return 0;
let g = Math.floor((n - 1) / 3 )+ 1;
return g*g + findCost(n - g);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He want to crush all of them. He feels that it would be boring to crush the candies randomly, so he devices a method to crush them. He divides these candies in minimum number of groups such than no group contains more than 3 candy. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much total cost would be incurred. Can you help him?1 <= N <= 10^9return the cost from the functionSample Input 1:
4
Explanation:
Query 1: First step John divides the candies in two groups of 3 and 1 candy respectively. Crushing one- one candy from both group would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int testCase = sc.nextInt();
while(testCase != 0){
int g= 0;
int n= sc.nextInt();
int temp = n;
while(temp > 0){
int t= temp/3;
int rem= temp - (t*3);
if(rem ==0){
temp=temp - t;
g= g + (t*t);
}
else{
temp = temp - (t+1);
g=g+(t+1)*(t+1);
}
}
System.out.println(g);
--testCase;
}
}
}, 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: 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: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: a, b, v = map(int, input().strip().split(" "))
c = abs(a-b)
t = c//v
print(t), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
System.out.print(Time(n,m,k));
}
static int Time(int A, int B, int S){
if(B>A){
return (B-A)/S;
}
return (A-B)/S;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the lexicographically smallest permutation P of integers from 1 to N such that:
1. It should not be strictly increasing i. e. there must be an i such that P[i] < P[i-1] for 2 <= i <= N
2. Sum of every three consecutive elements should be divisible by 3.The only line of input contains a single integer N, denoting the number of elements in the permutation.
3 <= N <= 15Print a single line containing N space separated integers denoting the elements of the permutation.Sample Input:
4
Sample Output:
1 3 2 4
Explanation:
Some permutations of 4 integers are:
1. [1, 2, 3, 4] - not valid since it is strictly increasing
2. [1, 3, 4, 2] - not valid since sum of [1, 3, 4] is not divisible by 3
3. [4, 3, 2, 1] - not valid because its not lexicographically smallest, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
int n = Integer.parseInt(br.readLine());
StringBuffer sb = new StringBuffer("");
if(n == 3)
System.out.println("1 3 2");
else if(n == 4)
System.out.println("1 3 2 4");
else
{
for(int i=1; i<n-3; i++)
{
sb.append(i+" ");
}
sb.append(n+" ").append(n-2+" ").append(n-1+" ").append(n-3+" ");
System.out.println(sb);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the lexicographically smallest permutation P of integers from 1 to N such that:
1. It should not be strictly increasing i. e. there must be an i such that P[i] < P[i-1] for 2 <= i <= N
2. Sum of every three consecutive elements should be divisible by 3.The only line of input contains a single integer N, denoting the number of elements in the permutation.
3 <= N <= 15Print a single line containing N space separated integers denoting the elements of the permutation.Sample Input:
4
Sample Output:
1 3 2 4
Explanation:
Some permutations of 4 integers are:
1. [1, 2, 3, 4] - not valid since it is strictly increasing
2. [1, 3, 4, 2] - not valid since sum of [1, 3, 4] is not divisible by 3
3. [4, 3, 2, 1] - not valid because its not lexicographically smallest, I have written this Solution Code: a=int(input())
if a==9:
print("1 2 3 4 5 9 7 8 6")
elif a==8:
print("1 2 3 4 8 6 7 5")
elif a==15:
print("1 2 3 4 5 6 7 8 9 10 11 15 13 14 12")
elif a==14:
print("1 2 3 4 5 6 7 8 9 10 14 12 13 11")
elif a==13:
print("1 2 3 4 5 6 7 8 9 13 11 12 10")
elif a==12:
print("1 2 3 4 5 6 7 8 12 10 11 9")
elif a==11:
print("1 2 3 4 5 6 7 11 9 10 8")
elif a==10:
print("1 2 3 4 5 6 10 8 9 7")
elif a==7:
print("1 2 3 7 5 6 4")
elif a==6:
print("1 2 6 4 5 3")
elif a==5:
print("1 5 3 4 2")
elif a==4:
print("1 3 2 4")
elif a==3:
print("1 3 2"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the lexicographically smallest permutation P of integers from 1 to N such that:
1. It should not be strictly increasing i. e. there must be an i such that P[i] < P[i-1] for 2 <= i <= N
2. Sum of every three consecutive elements should be divisible by 3.The only line of input contains a single integer N, denoting the number of elements in the permutation.
3 <= N <= 15Print a single line containing N space separated integers denoting the elements of the permutation.Sample Input:
4
Sample Output:
1 3 2 4
Explanation:
Some permutations of 4 integers are:
1. [1, 2, 3, 4] - not valid since it is strictly increasing
2. [1, 3, 4, 2] - not valid since sum of [1, 3, 4] is not divisible by 3
3. [4, 3, 2, 1] - not valid because its not lexicographically smallest, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n; cin >> n;
for(int i = 1; i <= n; i++)
a[i] = i;
do{
bool flag = 0;
for(int i = 2; i <= n; i++)
flag |= (a[i]-a[i-1] > 0 )?0:1;
for(int i = 1; i <= n-2; i++){
int sum = a[i] + a[i+1] + a[i+2];
if(sum%3 != 0)
flag = 0;
}
if(flag)
break;
}while(next_permutation(a+1, a+n+1));
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms:";
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, 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 = 2e5 + 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;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, 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(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
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;
t = sc.nextInt();
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 first term A and the common ratio R of a GP (Geometric Progression) and an integer N, your task is to calculate its Nth term.<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>NthGP()</b> that takes the integer A, R and N as parameter.
<b>Constraints</b>
1 <= A, R <= 100
1 <= N <= 10
Return the Nth term of the given GP.
<b>Note:</b> It is guaranteed that the Nth term of this GP will always fit in 10^9.Sample Input:
3 2
5
Sample Output:-
48
Sample Input:-
2 2
10
Sample Output:
1024
<b>Explanation:-</b>
For Test Case 1:- 3 6 12 24 48 96., I have written this Solution Code:
static int NthGP(int l, int b, int h){
int ans=l;
h--;
while(h-->0){
ans*=b;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where N is even. You have to pair up the elements into N/2 pairs such that each element is in exactly 1 pair. You need to find minimum possible X such that there exists a way to pair the N elements and for no pair sum of its elements is greater than X.First line contains N.
Second line contains N space separated integers, denoting array.
Constraints:
1 <= N <= 100000
1 <= elements of the array <= 1000000000Print a single integer, minimum possible X.Sample Input
4
3 1 1 4
Sample Output
5
Explanation: we can pair (1, 3) and (1, 4) so all pairs have sum less than or equal to 5., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int[]sort(int n, int a[]){
int i,key;
for(int j=1;j<n;j++){
key=a[j];
i=j-1;
while(i>=0 && a[i]>key){
a[i+1]=a[i];
i=i-1;
}
a[i+1]=key;
}
return a;
}
public static void main (String[] args) throws IOException {
InputStreamReader io = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(io);
int n = Integer.parseInt(br.readLine());
String str = br.readLine();
String stra[] = str.trim().split(" ");
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = Integer.parseInt(stra[i]);
}
a=sort(n,a);
int max=0;
for(int i=0;i<n;i++)
{
if(a[i]+a[n-i-1]>max)
{
max=a[i]+a[n-i-1];
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where N is even. You have to pair up the elements into N/2 pairs such that each element is in exactly 1 pair. You need to find minimum possible X such that there exists a way to pair the N elements and for no pair sum of its elements is greater than X.First line contains N.
Second line contains N space separated integers, denoting array.
Constraints:
1 <= N <= 100000
1 <= elements of the array <= 1000000000Print a single integer, minimum possible X.Sample Input
4
3 1 1 4
Sample Output
5
Explanation: we can pair (1, 3) and (1, 4) so all pairs have sum less than or equal to 5., I have written this Solution Code: n=int(input())
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
arr=sorted(arr,key=int)
start=0
end=n-1
ans=0
while(start<end):
ans=max(ans,arr[end]+arr[start])
start+=1
end-=1
print (ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where N is even. You have to pair up the elements into N/2 pairs such that each element is in exactly 1 pair. You need to find minimum possible X such that there exists a way to pair the N elements and for no pair sum of its elements is greater than X.First line contains N.
Second line contains N space separated integers, denoting array.
Constraints:
1 <= N <= 100000
1 <= elements of the array <= 1000000000Print a single integer, minimum possible X.Sample Input
4
3 1 1 4
Sample Output
5
Explanation: we can pair (1, 3) and (1, 4) so all pairs have sum less than or equal to 5., 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 a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
sort(a,a+n);
int ma=0;
for(int i=0;i<n;++i){
ma=max(ma,a[i]+a[n-i-1]);
}
cout<<ma;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton has N baskets containing apples. There are A<sub>i</sub> apples in ith basket. Newton wants to gift one apple to each of his M friends.
To do that he will select some baskets and gift all apples in these baskets to his friends. <b>Note</b> that he will gift apples if and only if there are exactly M apples in these subset of baskets.
Help him determine if it is possible to gift apple or not.The first line contains two space-separated integers – N and M.
The second line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub>.
<b> Constraints: </b>
1 ≤ N ≤ 100
1 ≤ M ≤ 10<sup>11</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>9</sup>
1 <= ∏ A<sub>i</sub> <= 10<sup>100</sup>Output "Yes" (without quotes) if Newton can distribute apples among his friends else "No" (without quotes).INPUT:
4 8
1 3 3 4
OUTPUT:
Yes
EXPLANATION:
Newton can use 1st, 2nd, 4th basket to give one apple to all his friends., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
const ll b = 10000;
ll pre[1000000];
int main(){
ll n,x;
cin >> n >> x;
vector<ll> a(n);
for(ll i=0 ; i<n ; i++){
cin >> a[i];
}
fill(pre, pre + 1000000, -1);
pre[0] = 0;
vector<pair<ll,ll>> v;
for(ll i=0 ; i<n ; i++){
if (a[i]>=b) v.push_back({ a[i],i });
else{
for(ll j=1000000-1 ; j>=0 ; j--) {
if (pre[j] < 0) continue;
if (j + a[i] < 1000000 && pre[j+a[i]] < 0) {
pre[j + a[i]] = j;
}
}
}
}
ll len = v.size();
for(ll i=0 ; i<(1 << len) ; i++){
ll sum = 0;
for(ll j=0 ; j<len ; j++) {
if (i & (1 << j)) sum += v[j].first;
}
ll d = x - sum;
if (d >= 0 && d < 1000000) {
if (pre[d] >= 0) {
cout << "Yes\n";
return 0;
}
}
}
cout << "No" << "\n";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import numpy as np
from collections import defaultdict
n=int(input())
a=np.array([input().strip().split()],int).flatten()
d=defaultdict(int)
for i in a:
d[i]+=1
d=sorted(d.items())
for i in d:
if(i[1]>1):
print(i[0],end=" ")
print(i[1])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int noOfElements = Integer.parseInt(in.readLine());
String [] elem = in.readLine().trim().split(" ");
int [] elements = new int [noOfElements];
for(int i=0 ; i< noOfElements; i++){
elements[i] = Integer.parseInt(elem[i]);
}
Arrays.sort(elements);
int count =0;
for(int i = 0 ; i < noOfElements-1 ; i++){
if(elements[i] == elements[i+1]){
count ++;
}
else if(count != 0){
System.out.print(elements[i]+" "+(count+1));
count =0;
System.out.println();
}
}
if(count != 0)
System.out.print(elements[noOfElements-1]+" "+(count+1));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n];
map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
m[a[i]]++;
}
for(auto it = m.begin();it!=m.end();it++){
if(it->second>1){
cout<<it->first<<" "<<it->second<<'\n';
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: def For_Loop(n):
string = ""
for i in range(1, n+1):
if i % 2 == 0:
string += "%s " % i
return string
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: public static void For_Loop(int n){
for(int i=2;i<=n;i+=2){
System.out.print(i+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 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 two integers ‘a’ and ‘m’. The task is to find modular multiplicative inverse of ‘a’ under modulo ‘m’.
Note: Print the smallest modular multiplicative inverse.First line of input contains a single integer T denoting number of test cases, next T lines contains two space separated integers depicting value of a and m.
Constraints :-
1 < = T < = 100
1 < = m < = 100
1 < = a < = mFor each test case, in a new line print the modular multiplicative inverse if exist else print -1.Sample Input:-
2
3 11
10 17
Sample Output:-
4
12, 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().trim());
while(t-->0){
String str[]=br.readLine().trim().split(" ");
int a=Integer.parseInt(str[0]);
int m=Integer.parseInt(str[1]);
int b;
int flag=0;
for(b=1; b<m; b++){
if(((a%m)*(b%m))%m == 1){
flag=1;
break;
}
}
if(flag==0){
System.out.println(-1);
}else{
System.out.println(b);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers ‘a’ and ‘m’. The task is to find modular multiplicative inverse of ‘a’ under modulo ‘m’.
Note: Print the smallest modular multiplicative inverse.First line of input contains a single integer T denoting number of test cases, next T lines contains two space separated integers depicting value of a and m.
Constraints :-
1 < = T < = 100
1 < = m < = 100
1 < = a < = mFor each test case, in a new line print the modular multiplicative inverse if exist else print -1.Sample Input:-
2
3 11
10 17
Sample Output:-
4
12, I have written this Solution Code: def modInverse(a, m):
for x in range(1, m):
if (((a%m)*(x%m) % m )== 1):
return x
return -1
t = int(input())
for i in range(t):
a ,m = map(int , input().split())
print(modInverse(a, m)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers ‘a’ and ‘m’. The task is to find modular multiplicative inverse of ‘a’ under modulo ‘m’.
Note: Print the smallest modular multiplicative inverse.First line of input contains a single integer T denoting number of test cases, next T lines contains two space separated integers depicting value of a and m.
Constraints :-
1 < = T < = 100
1 < = m < = 100
1 < = a < = mFor each test case, in a new line print the modular multiplicative inverse if exist else print -1.Sample Input:-
2
3 11
10 17
Sample Output:-
4
12, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int modInverseNaive(int a, int m) {
a %= m;
for (int x = 1; x < m; x++) {
if ((a*x) % m == 1) return x;
}
return -1;
}
signed main() {
int t;
cin >> t;
while (t-- > 0) {
int a,m;
cin >> a >> m;
cout << modInverseNaive(a, m) << '\n';
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b><em>Array List In Java</em></b>
There are n integers given.
<ul>
<li>Create an ArrayList</li>
<li>Add elements</li>
<li>Print the elements of arraylist.</li>
</ul>There is an integer n is given in first line of input.
In Second line, n space separated integer are given.
<b>Constraints</b>
1 <= n <= 10<sup>5</sup>
Print the elements of arraylist.Sample Input:
5
1 2 3 4 5
Sample Output:
1 2 3 4 5, I have written this Solution Code: // Java program for the above approach
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<>();
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=0;i<n;i++){
int a=sc.nextInt();
arr.add(a);
}
for(int i=0;i<n;i++){
System.out.print(arr.get(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, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
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>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
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>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code:
void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
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>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: public static void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
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>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: function patternMaking(N)
{
for(let j=1; j <= 2*N - 1; j++) {
let k;
if(j<= N) {
k = j;
} else {
k = 2*N - j;
}
let res = "";
for(let i=1; i<=2*k-1; i++) {
if(i<=k) {
res += i + " ";
} else {
res += 2*k - i + " ";
}
}
console.log(res);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
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>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: def pattern_making(n):
for i in range(1, n+1):
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=n-1
while i>=1 :
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=i-1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int f, s;
cin >> f >> s;
cout << (8*f)/s << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
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 java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int f = sc.nextInt();
int s = sc.nextInt();
int ans = (8*f)/s;
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code: a = list(map(int,input().strip().split()))[:2]
print(int((a[0]*8)/a[1])), 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){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 an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: def partition(array, low, high):
pivot = array[high]
i = low - 1
for j in range(low, high):
if array[j] <= pivot:
i = i + 1
(array[i], array[j]) = (array[j], array[i])
(array[i + 1], array[high]) = (array[high], array[i + 1])
return i + 1
def quick_sort(array, low, high):
if low < high:
pi = partition(array, low, high)
quick_sort(array, low, pi - 1)
quick_sort(array, pi + 1, high)
t=int(input())
for i in range(t):
n=int(input())
a=input().strip().split()
a=[int(i) for i in a]
quick_sort(a, 0, n - 1)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] quickSort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
public static int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: function quickSort(arr, low, high)
{
if(low < high)
{
let pi = partition(arr, low, high);
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
function partition(arr, low, high)
{
let pivot = arr[high];
let i = (low-1); // index of smaller element
for (let j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
let temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}, In this Programming Language: JavaScript, 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 |
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