Return to Question

I've done the given problem in roughly O(n√n).

I've done the given problem in roughly \$\mathcal{O}(n\sqrt{n})\$.

I've done the given problem in roughly O(n√n).

Similar problem: Sum Of Prime

I've done the given problem in roughly O(n√n).

#include <stdio.h>
int isPrime(long long int n)
{
 long long int i;
 for(i=2;i*i<=n;i++)
 {
 if(n%i==0)
 return 0;
 }
 return 1;
}
int main(void) {
 int i, t, prime;
 long long int n, start, end;
 scanf("%d", &t);
 for(i=0;i<t;i++)
 {
 prime = 0;
 scanf("%lld", &n);
 if(n<4)
 printf("No\n");
 else
 {
 if(n%2==0)
 printf("Yes\n");
 else
 {
 start = 2;
 end = n-2;
 while(start<=end)
 {
 if(isPrime(start) && isPrime(end))
 {
 printf("Yes\n");
 prime = 1;
 break;
 }
 else
 {
 start++;
 end–-;
 }
 }
 if(prime==0)
 printf("No\n");
 }
 }
 }
 return 0;
}

#include <stdio.h>
int isPrime(long long int n)
{
 long long int i;
 for(i=2;i*i<=n;i++)
 {
 if(n%i==0)
 return 0;
 }
 return 1;
}
int main(void) {
 int i, t, prime;
 long long int n, start, end;
 scanf("%d", &t);
 for(i=0;i<t;i++)
 {
 prime = 0;
 scanf("%lld", &n);
 if(n<4)
 printf("No\n");
 else
 {
 if(n%2==0)
 printf("Yes\n");
 else
 {
 start = 2;
 end = n-2;
 while(start<=end)
 {
 if(isPrime(start) && isPrime(end))
 {
 printf("Yes\n");
 prime = 1;
 break;
 }
 else
 {
 start++;
 end–-;
 }
 }
 if(prime==0)
 printf("No\n");
 }
 }
 }
 return 0;
}