POJ 3292 - Semi-prime H-numbers

算法

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发布日期: 2011-04-27

更新日期: 2026-11-06

文章字数: 701

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POJ 3292 - Semi-prime H-numbers
Time: 1000MS
Memory: 65536K
难度: 初级
分类: 同余模

0x00 POJ 原题

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

Sample Output

21 0
85 5
789 62

Source

Waterloo Local Contest, 2006.9.30

0x10 题目概述

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。

0x20 解题思路

感觉跟同余模扯不上关系。

筛法打表，再直接输出。

水题。

0x30 AC 源码

//Memory Time 
//4172K  63MS 

#include<iostream>
using namespace std;

const int size=1000001;

int H_Number[size+1];

/*筛法打表*/
void Table(void)
{
    memset(H_Number,0,sizeof(H_Number));  //H_Number[i]=0 表示 i为H-prime

    for(int i=5;i<=size;i+=4)
    {
        for(int j=5;j<=size;j+=4)
        {
            int multiply=i*j;
            if(multiply>size)
                break;

            if(H_Number[i]==0 && H_Number[j]==0)  //i与j均为H-prime
                H_Number[multiply]=1;  //multiply为H-semi-primes
            else
                H_Number[multiply]=-1; //multiply为H-composite
        }
    }

    int Pcount=0; //H-prime计数器
    for(int k=1;k<=size;k++)
    {
        if(H_Number[k]==1)
            Pcount++;
        H_Number[k]=Pcount;   //从1到k有Pcount个H-semi-primes
    }
    return;
}

int main(void)
{
    Table();
    int h;
    while(cin>>h && h)
        cout<<h<<' '<<H_Number[h]<<endl;

    return 0;
}