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): 7342    Accepted Submission(s): 4439

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

艾迪的兴趣是非常广泛的,最近他感兴趣的是素数.艾迪想找到所有可以被素数整除的数.但是他写不出这个程序。

can't write program, so Eddy has to ask intelligent you to help him,  

所以艾迪请求你帮他解决这个素数，

 he asks you to write a program which can do the number to divided into the multiply of prime number  factor .
他要求你把这个数分解成多个素数的积

 

Input

The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.

 输入将包含一个数字1 < x < = 65535每行代表的数量的元素集

Output

You have to print a line in the output for each entry with the answer to the previous question.

 每个题的答案在单独的一行输出

Sample Input

119412

 

Sample Output

112*2*13*181

 

Author

eddy

 

import java.util.Scanner;public class p1164 {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int[] primes = new int[7000];// 打表法primes[0] = 2;primes[1] = 3;primes[2] = 5;primes[3] = 7;int count = 4;for (int i = 11; i <= 65535; i += 2) {// 65535--素数个数为6542个boolean isPrime = true;for (int j = 0; primes[j] * primes[j] <= i && primes[j] != 0; j++) {if (i % primes[j] == 0) {isPrime = false;break;}}if (isPrime) {primes[count++] = i;}}String result = "";boolean isFirst;while (sc.hasNext()) {int x = sc.nextInt();result = "";isFirst = true;for (int i = 0; primes[i] <= x; i++) {if (x % primes[i] == 0) {if (isFirst) {result += primes[i];isFirst = false;} else {result += "*" + primes[i];}x /= primes[i];i = -1;}}System.out.println(result);}}}