poj3292

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

筛选法：p【i】 == 0表明4i+1为h-prime， p【i】 == 1表明4i+1为非h-semi-prime的h-composites，

               p【i】 == 2表明4i+1为h-semi-prime

如何筛选看代码

#include <iostream>
#include <cstring>
using namespace std;

#define MAXN 250005

int p[MAXN], sum[MAXN];

void init()
{
    int i, j;

    memset(p, 0, sizeof (p));
    memset(sum, 0, sizeof (sum));
   //(i<<2)+1 == 4*i+1
    for (i = 1; i <= 251; i++) {
        for (j = i; ; j++) {
            int tmp = ((i << 2) + 1) * ((j << 2) + 1); //tmp为两个h-number的乘积
            if (tmp > 1000001) break;
            tmp = (tmp - 1) >> 2;
            if (p[i] || p[j]) p[tmp] = 1; 
            else if (!p[i] && !p[j] && !p[tmp]) p[tmp] = 2;
        }
    }
    for (i = 1; i < MAXN; i++) //sum[i]存储前i个数中h-semi-prime的个数。
        if (p[i] == 2)
            sum[i] = sum[i-1] + 1;
        else sum[i] = sum[i-1];
}

int main()
{
    int n;

    init();
    while (cin >> n && n) {
         cout << n << " " << sum[(n-1)>>2] << endl;
    }
    return 0;
}