【hdu - 1164 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): 3554    Accepted Submission(s): 2182


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
 
 
 1 // Project name : 1164 ( Eddy's research I ) 
 2 // File name    : main.cpp
 3 // Author       : Izumu
 4 // Date & Time  : Sun Jul  8 21:07:17 2012
 5 
 6 
 7 #include <iostream>
 8 #include <stdio.h>
 9 #include <cmath>
10 using namespace std;
11 
12 int main()
13 {
14     int num;
15     while (cin >> num)
16     {
17         int* prime;
18         prime =new int[1000];
19         int top = -1;
20         int current_prime = 2;
21 
22         int CGetNextPrime(int);
23 
24         while (num != 1)
25         {
26             if (num % current_prime == 0)
27             {
28                 top++;
29                 prime[top] = current_prime;
30                 num /= current_prime;
31             }
32             else
33             {
34                 current_prime = CGetNextPrime(current_prime);
35             }
36         }
37 
38         // output 
39         for (int i = 0; i < top; i++)
40         {
41             cout << prime[i] << "*";
42         }
43         cout << prime[top] << endl;
44 
45     }
46     return 0;
47 }
48 /*********************************************************************/
49 bool CIsPrime(int k)
50 {
51     int stop = sqrt(k);
52     bool yes = true;
53     for (int i = 2; yes && i <= stop; i++)
54     {
55         if (k % i == 0)
56         {
57             yes = false;
58         }
59     }
60     return yes;
61 }
62 /*********************************************************************/
63 int CGetNextPrime(int num)
64 {
65     num++;
66     while (!CIsPrime(num))
67     {
68         num++;
69     }
70     return num;
71 }
72 // end 
73 // ism

 

posted @ 2012-07-08 21:17  ismdeep  阅读(179)  评论(0)  编辑  收藏  举报