hdu 5104 Primes Problem（素数判定）

Primes Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2838    Accepted Submission(s): 1275

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

题意：给一个n，问有多少种方式能找到三个非递减的素数之和等于n

思路：用素数筛选法记下所有的素数，然后两重循环枚举即可。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 2010int prime[N],vis[N*10];int cnt;void init(){    cnt=0;    memset(vis,0,sizeof(vis));    vis[0]=vis[1]=1;    for(int i=2;i<=10000;i++)    {        if(!vis[i])        {            prime[cnt++]=i;            for(int j=i*i;j<=10000;j+=i)                vis[j]=1;        }    }}int main(){    int n;    init();    while(~scanf("%d",&n))    {        long long ans=0;        for(int i=0;i<cnt&&prime[i]<=n;i++)        {            for(int j=i;j<cnt&&prime[i]+prime[j]<=n;j++)            {                int v=n-prime[i]-prime[j];                if(v>=prime[j]&&!vis[v])                    ans++;            }        }        printf("%lld\n",ans);    }    return 0;}