Semi-prime-H-numbers(POJ-3292)题解

╰+攻爆jí腚メ 2022-12-28 11:16 118阅读 0赞

### 文章目录 ###

*  题目分析
 *   *  输入格式
     *  输出格式
     *  输入
     *  输出
 *  算法分析
 *  解题标称

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# 题目分析 #

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.

## 输入格式 ##

> 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.

## 输出格式 ##

> 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.

## 输入 ##

21   
    85  
    789  
    0

## 输出 ##

21 0  
    85 5  
    789 62

# 算法分析 #

题目意思：  
H - numbers 是一个除以4余1的数  
H - primes是H - numbers 中的素数  
H-semi-primes是一个由两个H - primes组成的数字  
给定一个数h，从1-h个数中记录H-semi-primes的个数

这道题目直接将所有的H - primes 两两相乘进行组合，打表之后直接查表就可以得到最后结果

# 解题标称 #

#include <iostream>
    #include <cstdlib>
    #include <cstdio>
    using namespace std;
    
    const int MAXN=1000001+5;
    
    int vis[MAXN];  //记录数组中的下标是否为两个H-primes组成
    int ans[MAXN];  //记录数组到下标的所有数字中的H-semi-primes个数
    void Init()
    {
        
        memset(vis,0,sizeof(vis));
        //打表，在表中记录所有的H-semi-primes数字
        for(int i=5; i<=MAXN; i+=4)
            for(int j=5; j<=MAXN; j+=4)
            {
        
                int tmp=i*j;
                if(tmp>MAXN)
                    break;
                if(vis[i]==0&&vis[j]==0)
                    vis[tmp]=1;
                else
                    vis[tmp]=-1;
            }
        //打表，在表中记录所有的H-semi-primes数字
    
        //在ans表中记录从1到下标的范围内H-semi-primes的个数
        int cnt=0;
        for(int i=1; i<=MAXN; ++i)
        {
        
            if(vis[i]==1)
                cnt++;
            ans[i]=cnt;
        }
    }
    
    
    int main(int argc, const char * argv[]) {
        
        int n;
        
        Init();
        
        while(scanf("%d",&n)&&n){
        
            cout << n << " " << ans[n] << endl; //直接查询ans表即可
        }
        
        return 0;
    }