Primes Problem

Problem Description

Given a number n, please count how many tuple(p1, p2, p3) satisfied that p1<=p2<=p3, p1,p2,p3 are primes and p1 + p2 + p3 = n.

Input

Multiple test cases(less than 100), for each test case, the only line indicates the positive integer n(n≤10000).

Output

For each test case, print the number of ways.

Sample Input

Sample Output

#include <bits/stdc++.h>///Primes Problem
using namespace std;
#define N 10005
int prime[N];
int q[N];
int main()
{
    int i,j,k,n,l=0,sum;
    memset(q,0,sizeof(q));
    prime[0]=0;
    prime[1]=0;
    prime[2]=1;
    for(i=3;i<N;i++)//首先偶数一定不是素数,除了2
    {
        if(i%2)
            prime[i]=1;
        else
            prime[i]=0;
    }
    for(i=3;i<=sqrt(N);i++)
    {
        if(prime[i])
        for(j=i+i;j<N;j=j+i)//当i是素数的时候，i的所有的倍数必然是合数
        prime[j]=0;
    }
    for(k=0;k<10000;k++)
    {
        if(prime[k])
            q[l++]=k;
    }
    while(~scanf("%d",&n))
    {
        sum=0;
        for(int i=0;q[i]<n;i++)
            for(int j=i;q[j]<n-q[i];j++)
            {
                if(prime[n-q[i]-q[j]]&&n-q[i]-q[j]>=q[j])
                {
                    sum++;
                }
            }
        printf("%d\n",sum);
    }
    return 0;
}