1059. Prime Factors (25)

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此博客介绍了一个算法，用于找到给定正整数的所有质因数，并以特定格式输出。

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时间限制
50 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
HE, Qinming

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=2^2*11*17*101*1291

#include<iostream>
#include<math.h>
using namespace std;
typedef  long LL;
bool isprime(LL t){
    if(t==2) return true;
    for(int i=2;i<=sqrt(1.0*t);i++){
        if(t%i==0){
            return false;
        }
    }
    return true;
}
int main(){
    //freopen("in.txt","r",stdin);
    int n;
    scanf("%d",&n);
    printf("%d=",n);
    if(n==1){
        printf("%d",n);
        return 0;
    }
    while(!isprime(n)){
        for(int i=2;i<=sqrt(1.0*n);i++){
            if(isprime(i)&&n%i==0){
                int e=0;
                while(n%i==0){
                    n/=i;
                    e++;
                }
                if(n!=1){ 
                    if(e!=1) printf("%d^%d*",i,e);
                    else printf("%d*",i);
                }else  {
                    if(e!=1) printf("%d^%d",i,e);
                    else printf("%d",i);
                }
            }
        }
    }
    if(n!=1){
        printf("%d",n);
    }
    return 0;
}