1059 Prime Factors （25 分）

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p1^k1 * p2^k2 *…*pm^km, where pi’s are prime factors of N in increasing order, and the exponent ki is the number of pi – hence when there is only one pi, ki is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^211171011291

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思路：

• 从prime=2开始，判断是否能被n整除，如果可以的话一直除，直到无法整除，记录prime以及prime出现的次数，按格式输出
• 无法整除时,prime++;直到可以整除，循环此过程
  注意1=1的形式需要另外考虑

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代码：

#include<cstdio>
int main(){
	long long n,t=0,prime=2,count=0;
	scanf("%lld",&n);
	printf("%lld=",n);
	if(n==1) printf("1");
	while(n!=1){
		while(n%prime==0){//能除当前这个素数 
			n/=prime;
			count++;
		}	
		if(t!=0)  printf("*");
		if(count>1){
			printf("%lld^%lld",prime,count);
			t++;
		}else if(count==1){
			printf("%lld",prime);
			t++;
		}	
		//因为每个合数都可以由素数组成，因为前面已经尽可能多的除了素数，所以，得到的就是一个素数，不是合数
		//这里必须加n!=1，如果没有这个条件，当前面已经完全分解好质因数时，这里会陷入死循环
		while(n%prime!=0 &&n!=1)  prime++;
		count=0;
	}
	return 0;
}