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=;i<=n/i;i++){
         if(n%i==){
             int s =  ;
             while(n%i==){
                 s++ ;
                 n /= i ;
             }
             vc.push_back({i,s}) ;
         }
     }
     if(n>){
         vc.push_back({n,}) ;
     }
 }

 int main(){
     cin >> n ;

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

     return  ;
 }