POJ 3292

        这道题给出了H-numbers的定义——形如4*n+1的正整数，且题目里说只考虑这些数（For this problem we pretend that these are the only numbers.）。
        接着，题目里又说这些H-numbers可以分为三种：

1. unite：只有一个元素1
2. H-primes：只能由1*h构成
3. H-composites：其他情况。在这里也就是由其他H-numbers相乘得到，可能是两个相乘，也可能是三个或者更多相乘。由于题目里说这考虑H-numbers，所以不必考虑其他非H-numbers相乘得到H-numbers的情况。

        最后，题目里又提出H-composites的特例——H-semi-prime，仅由两个H-primes相乘得到的H-numbers。而题目就是要计算小于h的H-semi-prime个数。
        写这道题如果超时，但分析一下时间复杂度也能接受，那么就说明你的预处理还不够彻底，把所有能预先计算的都预先计算，包括H-semi-prime的数量，也就是答案也要先计算出来。

---

代码（G++）：

#include <iostream>
#include <cstring>
#include <vector>
#include <ctime>

#define MAX 1000002
using namespace std;

bool hprime[MAX], hnum[MAX];
int ans[MAX];
vector<int> hc, hp;

void initial()
{
    int c;
    long long t;

    memset(hprime, true, sizeof(hprime));
    hprime[0] = hprime[1] = false;
    for(int i=2; i<MAX; i++)
    {
        if(hprime[i])
        {
            if(i%4 == 1)
            {
                for(int j=i*2; j<MAX; j+=i)
                {
                    hprime[j] = false;
                }
            }else
                hprime[i] = false;
            if(hprime[i]) hp.push_back(i);
        }
    }

    memset(hnum, false, sizeof(hnum));
    for(size_t i=0; i<hp.size(); i++) if(hp[i] < MAX)
    {
        for(size_t j=i; j<hp.size(); j++)
        {
            t = (long long)hp[i]* hp[j];
            if(t <= MAX)
            {
                if(!hnum[t]) hnum[t] = true;
            }else{
                break;
            }

        }
    }
    c = 0;
    for(int i=1; i<MAX; i++)
    {
        if(hnum[i]) ++c;
        ans[i] = c;
    }
}

int main()
{
    int h;

    //double t1, t2;
    //t1=clock();
    initial();  //小于50ms
    //t2 = clock();
    //cout<<(t2-t1)/CLOCKS_PER_SEC*1000<<endl;

    while(cin>>h && h!=0)
    {
        cout<<h<<' '<<ans[h]<<endl;
    }
    return 0;
}

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题目：
Semi-prime H-numbers
Time Limit: 1000MS Memory Limit: 65536K

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