PAT1007 of prime conjecture

Of prime conjecture

Let d n is defined as: dn = p n + 1-pn, where p i is the i-th prime number. Clearly there is d 1 = 1, and for n> 1 there is an even number dn. "Primes to guess" that "there is an infinite number of adjacent and poor is a prime number 2."

Now given an arbitrary positive integer N (<10 ^{. 5} ), does not exceed Calculate number N satisfies the conjecture of primes.

Input formats:

It gives a positive integer in the input line N.

Output formats:

In the output line does not exceed the number N satisfy the conjecture of the primes.

Sample input:

20

Sample output:

4

Ideas:

3 and 5 are the first difference value is a prime number of from i = 3 starts verification i and i + 2 (less than N) whether, C language from 2 to root N, whether there is a prime number by N Analyzing the case of a number divisible is prime, the number of primes index records, the condition index ++, the final output value of index.

Code:

#include <stdio.h>
#include <math.h>
int judge(int n);
int main()
{
    int i,n,index;
    scanf("%d",&n);
    index=0;
    for(i=3;i<n;i++)
    {
        if(judge(i)&&judge(i+2)&&i<=n-2)
            index++;
    }
    printf("%d",index);
}
int judge(int n)
{
    int i;
    if(n==0||n==1)
        return 0;
    for(i=2;i<=sqrt(n);i++)
    {
        if(n%i==0)
            return 0;
    }
    return 1;
}