HDU 5104 Primes Problem（打表）

传送门： http://acm.hdu.edu.cn/showproblem.php?pid=5104

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
3
9

Sample Output
0
2

把我wa哭了 判断结果那 写成prime[j]&&prime[i] 就时间超限
prime[i]&&prime[j] 就过了 不是很懂

code：

#include<cstdio>
#include<iostream>
//#include<bits/stdc++.h>
#include <string.h>
using namespace std;

int is_prime(int n )
{

    for(int i=2 ;i*i<=n;i++ )
    {
        if ( n % i ==0)
            return 0;
    }
    return 1;
}
int main()
{
    int prime[10005];
    memset(prime, 0 ,sizeof(prime));
//    for (int i=2; i<=10002; i++)
//    {
//        int flag=0;
//        for (int j=2; j*j<=i;j++)
//        {
//            if (i%j==0)
//            {
//                flag=1;
//                break;
//            }
//
//        }
//        if (flag==0)
//            prime[i]=1;
//
//    }
for(int i=2 ;i<=10002;i++)
    {
        if ( is_prime(i) )
            prime[i] = 1;
    }
    int n;

    while (cin>>n)
    {
        int num=0;
        for (int i=2; i<=n; i++)
        {
            for (int j=i; j<=(n-i)/2; j++)
            {
                if (prime[i]&&prime[j]&&prime[n-i-j])
                {
                    num++;
                }

            }
        }
        cout<<num<<endl;
    }



    return 0;
}