HDU 1164 Eddy's research I（将数分解成素数之积）

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该博客介绍了HDU 1164题目——Eddy's research I，主要讨论如何将非素数分解为素数的乘积。提供了两种解决方案，包括暴力求解和使用素数表优化的方法。

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Eddy's research I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8298 Accepted Submission(s): 5085

Problem Description

Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .

Input

The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.

Output

You have to print a line in the output for each entry with the answer to the previous question.

Sample Input

  
   11
9412

Sample Output

  
   11
2*2*13*181

Author

eddy

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1164

题意：将一个非素数分解成素数相乘的形式，若是素数，直接输出。

AC代码1：(暴力)

#include<stdio.h>
int main()
{
    int i,n;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=2;i*i<=n;i++)
        {
            if(n%i==0)
            {
                n=n/i;
                printf("%d*",i);
                i=1;
            }
        }
        printf("%d\n",n);
    }
    return 0;
}

AC代码二：(打表)

#include <iostream>
#include <cstring>
using namespace std;
#define maxn 100000
int a[maxn];
bool prime[maxn];
void getPrime()
{
    memset(prime,true,sizeof(prime));
    for(int i=2; i<maxn; i++)
    {
        if(prime[i]==true)
        {
            for(int j=i+i; j<maxn; j+=i)
                prime[j]=false;
        }
    }
}
int main()
{
    int x;
    getPrime();
    while(cin>>x)
    {
        if(prime[x])
        {
            cout<<x<<endl;
            continue;
        }
        for(int i=2; i*i<=x; i++)
        {
            if(prime[i]&&(x%i==0))
            {
                x/=i;
                cout<<i<<"*";
                i=1;
            }
        }
        cout<<x<<endl;
    }
    return 0;
}