PAT1059Prime Factors

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 = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​.

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 = p​1​​^k​1​​*p​2​​^k​2​​*…*p​m​​^k​m​​, where p​i​​'s are prime factors of N in increasing order, and the exponent k​i​​ is the number of p​i​​ -- hence when there is only one p​i​​, k​i​​ is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291
思路：算术基本定理（唯一分解定理）
用从小到大的素数整除n，vector保留素数及其对应的指数

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std ;

typedef pair<int,int> PII ;
vector<PII> vc ;
int n ;

void get(){
    for(int i=2;i<=n/i;i++){
        if(n%i==0){
            int s = 0 ;
            while(n%i==0){
                s++ ;
                n /= i ;
            }
            vc.push_back({i,s}) ;    
        }
    }
    if(n>1){
        vc.push_back({n,1}) ;
    }
}

int main(){
    cin >> n ;
    
    if(n==1){
        printf("1=1") ;
    }else{
        printf("%d=",n) ;
        get() ;
        int la = vc.size() ;
        for(int i=0;i<la;i++){
            if(i){
                printf("*") ;
            }
            printf("%d",vc[i].first) ;
            if(vc[i].second>1){
                printf("^%d",vc[i].second) ;
            }
        }
    }
    
    return 0 ;
}