[poj 3292] Semi-prime H-numbers 筛数

Semi-prime H-numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8514 Accepted: 3717

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

Source
Waterloo Local Contest, 2006.9.30

题目链接:http://poj.org/problem?id=3292
题意:定义一种数叫H-numbers,它是所有能除以四余一的数。

在H-numbers中分三种数:

1、H-primes,这种数只能被1和它本身整除,不能被其他的H-number整除,例如9是一个H-number,能被1,3,9整除,但3不是H-number,所以他是H-primes。

2、H-semi-primes是由两个H-primes相乘得出的。

3、剩下的是H-composite。

问给一个数,求1到这个数之间有多少个H-semi-primes。

思路:枚举mod4=1的数作筛数,3个标记;

代码

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
#define maxn 1000005
int p[maxn];
int n;
int ans[maxn];
void Init()
{

    for(int i=5;i<=maxn;i+=4)
    for(int j=5;j<=maxn;j+=4)
    {

        if(i*j>maxn) break;
        if(!p[i]&&!p[j]) p[i*j]=1;
        else p[i*j]=2;
    }
    int ret=0;

    for(int i=1;i<=maxn;i++)
    {
        if(p[i]==1) ret++;
        ans[i]=ret;

    }
}


int main()
{
    Init();
    while(scanf("%d",&n))
    {
        if(!n) return 0;
        printf("%d %d\n",n,ans[n]);
    }

}

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