HDUOJ----Eddy's research I

最新推荐文章于 2019-03-10 15:53:34 发布

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最新推荐文章于 2019-03-10 15:53:34 发布

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文章标签： java

原文链接：http://www.cnblogs.com/gongxijun/p/3245440.html

本文介绍了一种质因数分解算法，并提供了两种实现方式。该算法可以将任意整数分解为质因数的乘积形式，对于理解数学中的质数概念及编程实现具有一定的参考价值。

Eddy's research I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5050 Accepted Submission(s): 3027

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

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两种做法：

代码：

 1 #include<iostream>
 2 using namespace std;
 3 const int maxn=65540;
 4 int arr[maxn]={2,3,5,7};
 5 void  prime()
 6 {
 7     bool flag;
 8     int k=4;
 9     for(int i=11;i<maxn;i++)
10     {
11         flag=true;
12      for(int j=2;j*j<=i;j++)
13      {
14          if(i%j==0)
15          {
16            flag=false;
17          }
18      }
19      if(flag)arr[k++]=i;
20     }
21 }
22 int main()
23 {
24   int n,count;
25   prime();
26  while(cin>>n)
27  {
28      count=0;
29  while(n!=1)
30  {
31     for(int i=0;n!=1;i++)
32     { 
33       if(n%arr[i]==0)
34       {
35           if(count++==0)
36           cout<<arr[i];
37           else
38             cout<<"*"<<arr[i];
39           n/=arr[i];
40           break;
41       }
42     }
43  }
44  cout<<endl;
45  }
46 return 0;
47 }

View Code

代码：

 1 #include<stdio.h>
 2 void show(int a)
 3 {
 4     int i=1,n=0;
 5     while(i<=a&&a!=1)
 6     {
 7         i++;
 8        while(a%i==0)
 9        {
10           a/=i;  n+=1;
11           printf(n==1?"%d":"*%d",i);
12        }
13 
14     }
15     puts("");
16 
17 }
18 int main()
19 {
20     int x;
21  while(scanf("%d",&x)!=EOF)
22  show(x);
23  return 0 ;
24 }