hdu 1164 Eddy's research I （数论）_eddy是个acmer他不仅喜欢做acm题而且对于纸牌-优快云博客

本文链接：https://blog.youkuaiyun.com/nealgavin/article/details/7460166

本文介绍了一个质因数分解的程序实现方法，通过判断素数并将输入的整数分解为若干个质数的乘积形式。该程序使用了C++语言进行编写，并提供了完整的源代码。

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Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 2 Accepted Submission(s) : 2

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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

#include<iostream>
#include<cstring>
#define N 68535
#define M 300
using namespace std;
bool isnop[N];//判断素数
int prime[N];//按顺序存素数
int ans;
void isprime()
{
   ans=-1;
  memset(isnop,0,sizeof(isnop));
  isnop[0]=isnop[1]=1;
  for(int i=2;i<M;i++)
  {
    if(!isnop[i])
    {
      prime[++ans]=i;
      for(int j=i+i;j<N;j+=i)
      isnop[j]=1;
    }
  }
  for(int i=M-1;i<N;i++)
  if(!isnop[i])
  prime[++ans]=i;
}
int main()
{
  int m,i;
  isprime();
  while(cin>>m)
  {
    i=0;
    while(1)
    {
      if(m%prime[i]==0&&m)
      {
        m/=prime[i];
        if(m!=1)
        cout<<prime[i]<<"*";
        else
        {
          cout<<prime[i]<<endl;
          break;
        }
      }
      else ++i;
    }
  }
}