hdoj-1164-Eddy‘s research I【分解质因数】

最新推荐文章于 2019-03-10 15:53:34 发布

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本文介绍了一种用于将任意正整数分解为质因数的算法，详细解释了算法的实现过程，并通过实例展示了如何将给定的整数分解为最小质因数的乘积。

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Eddy's research I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7537 Accepted Submission(s): 4579

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

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#include<stdio.h>
int main(){
	int  x,i;
	while(~scanf("%d",&x)){
		i=2;
		while(1){
		    if(i==x) {
				printf("%d\n",i);
				break;
			}
			if(x%i==0) {
				printf("%d*",i);
				x/=i;
			}
			else {
				++i;  
			}
			
		}
	}
	return 0;
}

刚开始不明白为什么可以直接进行++i，而不再考虑 i 是否为质数,之后才明白过来：因为从2开始分解，一直分解到剩下的因子不再能够被2整除为止，才进行对下一个数的分解，而对于合数，它总能够分解为几个素数的乘积；然而此合数的素数因子已经在之前都被完全提取出来了，所以对于x 是合数的情况，不可能再找到 x%i==0 （注意：此处的X已经不是最开始输入的那个x了），另外：每个素数之间又没有什么关系，先去分解哪个质数，其最终得到的质因子是完全一样的，所以也不用担心前面的分解结果会对后面的分解造成影响！！！