HDUOJ----Eddy's research I

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

 

 

Recommend

JGShining

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 }

View Code