1059. Prime Factors (25)

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 *…*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 = p₁^k₁ * p₂^k₂ *…*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

 #include<stdio.h>
 #include<math.h>
 #include<vector>
 using namespace std;
 int main()
 {
     int n,i,j;
     vector<int> vv;
     for(i= ;i<=;i++)
     {
         bool is = true;
         for(j=;j<=sqrt(i*1.0);j++)
         {
             if(i % j == )
             {
                 is = false;
                 break;
             }
         }
         if(is) vv.push_back(i);
     }
     while(scanf("%d",&n)!=EOF)
     {
         printf("%d=",n);
         bool fir = true;
         if(n == ) printf("");
         else
         {
             i = ;
             while(n != )
             {
                 if(n % vv[i] == )
                 {
                     int tem = ;
                     while(n % vv[i] == )
                     {
                         ++tem;
                         n = n / vv[i];
                     }

                     if(fir)
                     {
                         printf("%d",vv[i]);
                         fir = false;
                     }
                     else
                     {
                         printf("*%d",vv[i]);
                     }

                     if(tem > )
                     {
                         printf("^%d",tem);
                     }
                 }

                 ++i;
             }
         }

         printf("\n");
     }
     return ;
 }