hdu 1164 Eddy's research I【快速打印素数表】【水题】

Eddy's research I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8532    Accepted Submission(s): 5243

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

 

思路：把素数打出来，然后存入一个数组中，循环处理n即可。

AC代码：

#include<stdio.h>#include<string.h>#include<math.h>using namespace std;#define maxn 65550int Is_or[maxn];int su[maxn];int cont;void init(){        memset(Is_or,0,sizeof(Is_or));    for(int j=2;j<sqrt(maxn);j++)//    {        if(Is_or[j]==0)//去掉合数的倍数.        for(int k=j+j;k<=maxn;k+=j)//去掉倍数.(把这么些个合数的倍数都标记上这个数不是素数.)        Is_or[k]=1;    }    for(int i=2;i<=maxn;i++)    {        if(Is_or[i]==0)        {            su[cont++]=i;        }    }}int main(){    cont=0;    init();    int n;    while(~scanf("%d",&n))    {        int f=0;        while(1)        {            if(n==1)break;            for(int i=0;i<cont;i++)            {                if(n%su[i]==0)                {                    n/=su[i];                    if(f==0)                    printf("%d",su[i]),f++;                    else printf("*%d",su[i]);                    break;                }            }            if(n==1)break;        }        printf("\n");    }}