Eddy's research I

Eddy's research I

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

Total Submission(s) : 14   Accepted Submission(s) : 12

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

输入一个数，输出相乘能够等于该数的素数；

#include <stdio.h>
#define Max(a,b) a>b? a:b
#define Min(a,b) a>b? b:a
#include <string.h>
int main()
{
    int i,j,a[66535]={0},NUM,b[10000],k;
    for(i=2;i<=65535;i++)
        a[i]=1;
    for(i=2;i<=65535;i++)   /*素数筛选出该范围内所以素数*/
        if(a[i])
            for(j=i+i;j<=65535;j+=i)
                a[j]=0;
    while(scanf("%d",&NUM)!=EOF)
    {
        if(a[NUM])          /*如果该数字就是素数，那么只需要输出该数字就是答案*/
        {
            printf("%d\n",NUM);
            continue;
        }
        k=0;                    
        for(i=NUM-1;i>=2;i--)    /*从该数组的前一位，一直向前查找*/
            while(a[i]&&NUM%i==0)  /*如果查找到能够被该数整除的数组，存取，并且更新该数*/
            {
                NUM/=i;
                b[k++]=i;
            }
        for(j=k-1;j>=0;j--)
        {
            printf("%d",b[j]);      /*反序输出即为所求答案*/
            if(j!=0)putchar('*');
        }
        putchar(10);
    }
    return 0;
}

 修改：2015.6.2

根据所求的，只需要知道能够被该数整除的素数，依次取出那些数以及个数即可、

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int Pr[65540]={0};
void Cread()
{
    int i,j;
    for(i=1;i<=65535;i++)Pr[i]=1;
    for(i=2;i<=65535;i++)
        if(Pr[i])
            for(j=2;i*j<=65535;j++)
                Pr[i*j]=0;
    return ;
}
int main()
{
    int N,i,j,k,sign=0;
    Cread();
    while(scanf("%d",&N)!=EOF)
    {
        sign=0;
        for(i=2;i<=N;i++)
        {
            if(N==1)break;
            if(Pr[i])
            {
                j=N;
                while(N%i==0)
                {
                    if(sign)putchar('*');
                    printf("%d",i);
                    sign=1;
                    N/=i;
                }
            }
        }putchar(10);
    }
    return 0;
}