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 = ;
 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 ;
 }