poj-3292-Semi-prime H-numbers

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
一个H-number是所有的模四余一的数。

如果一个H-number是H-primes 当且仅当它的因数只有1和它本身（除1外）。

一个H-number是H-semi-prime当且仅当它只由两个H-primes的乘积表示。

H-number剩下其他的数均为H-composite。

给你一个数h,问1到h有多少个H-semi-prime数。
直接暴力打表就好了啊

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

const int maxn=1000001;
int num[maxn];

void init()
{
    for(int i=5;i<=maxn;i+=4)
        for(int j=5;j<=maxn;j+=4)
    {
        int mul=i*j;
        if(mul>maxn) break;
        if(num[i]==0&&num[j]==0)
            num[mul]=1;
        else num[mul]=-1;
    }
    int cnt=0;
    for(int k=1;k<=maxn;k++)
    {
        if(num[k]==1) cnt++;
        num[k]=cnt;
    }
}

int main()
{
    int n;
    init();
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        printf("%d %d\n",n,num[n]);
    }
    return 0;
}