Given an integer $N$, determine if it is a prime a number. A number is called prime if it is only di…
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pretests are so weak…
What’s the problem with this code? It gets wrong ans on test 6
N = int(input())
def isprime(n):
if (n > 1):
for i in range(2, n):
if n % i == 0:
return False
else:
return True
else:
return False
if isprime(N):
print('Yes')
else:
print('No')
@CSBMC_A You are returning True as soon as you find an i which cannot divide n without a remainder.
The loop should look something like this:
for i in range(2, n):
if n % i == 0:
return False
return True
#include <stdio.h>
int main() {
int n,c=0,i;
for(i=2;i<n;i++)
{ if(n%i==0)
c++;
break;
}
if(c==0)
{ printf(“Yes”);
}
else
{ printf(“No”);
}
return 0;
}
whats the problem ?
1 is not a prime number.your code takes 1 as prime
no;
it gets wrong answer in last test case
you have not even taking the inputs.
Your code is just printing YES maybe.
And the last TC might be NO.
According to your definition, the only prime number is 1 (which is ironic, since 1 is not actually prime). All other positive integers are divisible by (at least) themself and 1, making them non-prime according to your definition.
Agree with you. Using Fermat’s Little theorem i got accepted (Was not expecting though used just for testing purpose).
#include <stdio.h>
int main()
{
int n,i,flag=0;
scanf("%d",&n);
if(n==1||n==0){
printf("NO");
}
else{
for(i=2;i<=n/2;i++){
if(n%i==0)
flag=1;
}
if(flag==0)
printf("YES");
else
printf("NO");
}
return 0;
}
Please help me
What is wrong in this
Hint: Every non-prime is divisible by 2 and/or 3… if a number is not divisible by either, the number is prime.
What’s the problem here?
#include <stdio.h>
int main()
{
int n;
int check = 0;
printf("Enter a number:");
scanf("%d", &n);
for(int i = 2; i < n; i++){
if(n % i == 0){
check = 1;
break;
}
}
if(check == 0 && n != 1){
printf("Yes");
}
else{
printf("No");
}
return 0;
}
Remove extra output : “printf(“Enter a number:”);”
what is wrong in this
#include<iostream>
using namespace std;
int main(){
int N;
cin >> N;
if(N%2!=0)
{
cout << "Yes" <<endl;
}
else
{
cout << "No" <<endl;
}
return 0;
}
@Abdhulla.985108 You are checking if the code is odd or even. You are not checking if it is prime.
#include <stdio.h>
int main ()
{
int N, i, k=0;
scanf("%d", &N);
if (N==1000)
if ( N==1 || N==0)
k=1;
for(i=2; i<N; i++)
{
if(N%i==0)
k++;
break;
}
if(k==0)
printf("\Yes");
else
printf("\No");
return 0;
}
what is the problem, please help me