hdu_5104 Primes Problem()

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5104

rimes Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2844    Accepted Submission(s): 1277

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

 

题意：输入一个数字n，找出三个数字p1,p2,p3，满足p1<=p2<=p3并且p1+p2+p3=n

//筛素数
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 10005;
bool pri[N];
int prime[N];
int cnt;
void init()
{
    cnt = 0;
    pri[0] = pri[1] = 1;
    for(int i = 2; i < N; i++)
    {
        if(!pri[i]){
            prime[cnt++] = i;
            for(int j = i+i; j < N; j+=i)
            {
                pri[j] = 1;
            }
        }
    }
    return;
}
int main()
{
    int n;
    init();
    while(~scanf("%d",&n))
    {
        int ans = 0;

        //printf("%d\n",cnt);
        for(int i = 0; i < cnt; i++)
        {
            if(3*prime[i]>n) break;
            for(int j = i; j < cnt; j++)
            {
                if(prime[i]+2*prime[j]>n) break;
                //for(int k = j; k < cnt; k++)
                //{
                //    if(prime[i]+prime[j]+prime[k]==n) ans++;// printf("%d %d %d\n",prime[i],prime[j],prime[k]);printf("%d %d %d\n",i,j,k);}
               // }
               if(!pri[n-prime[i]-prime[j]]) ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}