POJ-3292-Semi-prime H-numbers-数筛法

http://poj.org/problem?id=3292

按素筛法筛出1e6的 H-prime-number

再存起来，暴力求出所有Semi-prime H-numbers，存起来

每次二分查找即可

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>#include<stack>using namespace std;  int hprim[1000000];bool f[1000000+500];int h_s_p[120000+500]; int main(){__int64 i,j;f[1]=false;for (i=4;i<=1000000;i+=4)//筛H-prime-number{__int64 re=i+1;if (f[re]==false){for (j=re*2;j<=1000000;j+=re)f[j]=true;}}int ok=0;for (i=4;i<=1000000;i+=4)//存储h-prime-number{if (f[i+1]==false)hprim[++ok]=i+1;}int cun=0;for (i=1;i<=ok;i++)//构造H--semi-primes{for (j=i;j<=ok;j++){__int64 ret=hprim[i];ret*=hprim[j];if (ret>1000001)break; h_s_p[++cun]=ret;}}sort(h_s_p+1,h_s_p+1+cun);cun=unique(h_s_p+1,h_s_p+cun)-h_s_p; //去重 int n;while(scanf("%d",&n)!=EOF){if (!n)break;int it=upper_bound(h_s_p+1,h_s_p+1+cun,n)-h_s_p;it--;printf("%d %d\n",n,it); }return 0;}