PAT_A1059#Prime Factors

Source:

PAT A1059 Prime Factors (25 分)

Description:

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

Keys:

• 模拟题

Attention:

• 注意N==1的情况

Code:

 /*
 Data: 2019-07-08 20:14:37
 Problem: PAT_A1059#Prime Factors
 AC: 14:40

 题目大意：
 打印正整数N的质因子，及其个数
 */
 #include<cstdio>
 #include<vector>
 #include<cmath>
 using namespace std;
 typedef long long LL;
 struct node
 {
     LL p,k;
 };

 int main()
 {
     LL n;
     vector<node> fac;
     scanf("%lld", &n);
     printf("%lld=", n);
     for(LL i=; i<=(LL)sqrt(1.0*n); i++)
     {
         if(n%i == )
         {
             node t = node{i,};
             while(n%i== && n!=)
             {
                 t.k++;
                 n /= i;
             }
             fac.push_back(t);
         }
         if(n == )
             break;
     }
     if(n != )
         fac.push_back(node{n,});
     if(fac.size()==)
         printf("");
     for(int i=; i<fac.size(); i++)
     {
         printf("%lld", fac[i].p);
         if(fac[i].k != )
             printf("^%lld", fac[i].k);
         printf("%c", i==fac.size()-?'\n':'*');
     }

     return ;
 }