UVA 11105 Semiprime Hnumbers（H半素数）

题意：所有形如4n+1(n为非负整数)的数叫H数。定义1是唯一的单位H数，H素数是指本身不是1，且不能写成两个不是1的H数的乘积。H-半素数是指能写成两个H素数的乘积的H数(这两个数可以相同也可以不同)。输入一个H数h(h <=1000001)，输出1~h之间有多少个H-半素数。

分析：

1、筛选法求H素数。

2、再枚举求H-半素数。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e6 + 10;
const int MAXT = 10000 + 10;
using namespace std;
vector<int> Hprime;
int vis[MAXN];
void init(){
    for(int i = 5; i < MAXN; i += 4){
        if(vis[i]) continue;
        Hprime.push_back(i);
        for(int j = i * 2; j < MAXN; j += i){
            vis[j] = 1;
        }
    }
}
int solve(int n){
    int ans = 0;
    memset(vis, 0, sizeof vis);
    int len = Hprime.size();
    for(int i = 0; i < len; ++i){
        if(Hprime[i] >= n) break;
        if((LL)Hprime[i] * Hprime[i] > n) break;
        for(int j = i; j < len; ++j){
            if(Hprime[j] >= n) break;
            LL tmp = (LL)Hprime[i] * Hprime[j];
            if(tmp > n) break;
            if(vis[tmp]) continue;
            ++ans;
            vis[tmp] = 1;
        }
    }
    return ans;
}
int main(){
    init();
    int n;
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        printf("%d %d\n", n, solve(n));
    }
    return 0;
}