hdu 1164:Eddy's research I（水题，数学题，筛法）

Eddy's research I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5793    Accepted Submission(s): 3459

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   |   We have carefully selected several similar problems for you:  1215 1286 1211 1299 1163 

---

 

　　水题。

　　用筛法求一个数的质因子相乘的形式。

　　筛法的“筛子”是逐步完善的，可以节省时间。

　　代码：

 #include <iostream>
 
 using namespace std;
 bool prime(int n)   //判断一个数是不是素数
 {
     for(int i=;i<n;i++)
         if(n%i==)
             return ;
     return ;
 }
 bool a[];  //存储素数
 int b[];
 int main()
 {
     int x;
     while(cin>>x){
         int n= ;
         int num = x;
         int i=;
         while(num!=){
             if(num%i==){   //x被i整除
                 if(a[i]){    //i是素数
                     b[++n] = i;
                     num/=i;
                     i=;
                 }
                 else if(prime(i)){
                     a[i] = true;
                     b[++n] = i;
                     num/=i;
                     i=;
                 }
             }
             i++;
         }
         for(i=;i<n;i++)
             cout<<b[i]<<'*';
         cout<<b[i]<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013