Primes Problem （HDU5104） 素数问题

Primes Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 796    Accepted Submission(s): 368

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

39

 

Sample Output

02

 

找到小于10000的素数，用素数筛选法。

避开三重循环，采用二重循环，如果存在p3满足条件，则必定会有p3 = n - p1 - p2，如果p3是素数并且p3大于等于p2，则计数器加一。

因为使用三重循环直接查找，到时TLE了一回。。。。。。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int su[10000] = {2,3,5};int prim[10000];int init(){    memset(prim,0,sizeof(prim));    prim[2] = 1;    prim[3] = 1;    prim[5] = 1;    int i,j,k,m = 2,bj;    k = 3;    for(i = 7; i <= 10000; i = i + m)    {        bj = 1;        m = 6 - m;        for(j = 0; su[j]*su[j] <= i; j++)        {            if(i%su[j] == 0)            {                bj = 0;                break;            }        }        if(bj == 1)        {            su[k++] = i;            prim[i] = 1;        }    }    return k;}int main(){    int t,i,j,m;    int k = init();    while(~scanf("%d",&t))    {        int num = 0;        for(i = 0; i < k; i++)        {            for(j = i; j < k; j++)            {                m = t - su[i] - su[j];                if(m < 0)                {                    break;                }                if(prim[m] == 1 && m >= su[j])                    num++;            }        }        printf("%d\n",num);    }    return 0;}