poj3292 Semi-prime H-numbers

题意：

Description

This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Input

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

Output

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

Sample Input

21 
85
789
0

Sample Output

21 0
85 5
789 62

思路：

先打H-素数表，再打只有两个素因子的表，最后二分。

实现：

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN = 1000001;
bool is_prime[MAXN + 5];
int x;
vector<int> res;
void init()
{
    vector<int> prime;
    memset(is_prime, 1, sizeof(is_prime));
    for (int i = 5; i <= MAXN; i += 4)
    {
        if (is_prime[i])
        {
            prime.push_back(i);
            for (int j = 5; i * j <= MAXN; j += 4)
            {
                is_prime[i * j] = false;
            }
        }
    }
    for (int i = 25; i <= MAXN; i += 4)
    {
        if (is_prime[i])
            continue;
        int tmp = i;
        int cnt = 0;
        for (int j = 0; prime[j] * prime[j] <= tmp; j++)
        {
            if (cnt > 2)
                break;
            while (tmp % prime[j] == 0)
            {
                cnt++;
                tmp /= prime[j];
            }
        }
        if (tmp != 1)
            cnt++;
        if (cnt <= 2)
            res.push_back(i);
    }
}

int main()
{
    init();
    while (scanf("%d", &x), x)
    {
        printf("%d %d
", x, upper_bound(res.begin(), res.end(), x) - res.begin());
    }
    return 0;
}