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
9
Sample Output
0
2
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 = ;
bool pri[N];
int prime[N];
int cnt;
void init()
{
cnt = ;
pri[] = pri[] = ;
for(int i = ; i < N; i++)
{
if(!pri[i]){
prime[cnt++] = i;
for(int j = i+i; j < N; j+=i)
{
pri[j] = ;
}
}
}
return;
}
int main()
{
int n;
init();
while(~scanf("%d",&n))
{
int ans = ; //printf("%d\n",cnt);
for(int i = ; i < cnt; i++)
{
if(*prime[i]>n) break;
for(int j = i; j < cnt; j++)
{
if(prime[i]+*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 ;
}