POJ 3292, Semi-prime H-numbers

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最新推荐文章于 2023-08-18 09:01:00 发布

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原文链接：http://www.cnblogs.com/asuran/archive/2009/10/22/1587802.html

本文探讨了一种特殊的数论问题，即H数及其H-semi-primes的概念。H数定义为正整数中比四的倍数大一的数，如1,5,9等。H-semi-primes是指由两个H素数相乘得到的H数。文章通过编程实现了一个算法，用于计算指定范围内H-semi-primes的数量。

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Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4166 Accepted: 1652

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

// POJ3292.cpp : Defines the entry point for the console application.
//

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

int main(int argc, char* argv[])
{
    unsigned long long const SIZE = 1000005;
    unsigned long long MAX = unsigned long (sqrt(SIZE*1.0));
    int H[SIZE];
    memset(H, 0, sizeof(H));
    unsigned long long num;
    for (unsigned long long i = 5; i < MAX; i+=4)
    {
        for(unsigned long long j = i; (num = i * j )< SIZE; j+=4)
        {
            if (H[i] != 0 || H[j] != 0)
                H[num] = -1;
            else if (H[num]!=-1)
                H[num] = 1;
        }
    }

    for (int i = 1; i < SIZE; ++i)
    {
        if (H[i] == 1) H[i] = H[i-1] + 1;
        else H[i] = H[i-1];
    }

    int N;
    while(scanf("%d", &N)!=EOF && N!=0)
        cout << N << " " << H[N] << endl;

    return 0;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/22/1587802.html