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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:97532468Sample Output:
97532468=2^2*11*17*101*1291
1 #include<stdio.h> 2 #include<math.h> 3 #include<vector> 4 using namespace std; 5 int main() 6 { 7 int n,i,j; 8 vector<int> vv; 9 for(i=2 ;i<=10100;i++) 10 { 11 bool is = true; 12 for(j=2;j<=sqrt(i*1.0);j++) 13 { 14 if(i % j == 0) 15 { 16 is = false; 17 break; 18 } 19 } 20 if(is) vv.push_back(i); 21 } 22 while(scanf("%d",&n)!=EOF) 23 { 24 printf("%d=",n); 25 bool fir = true; 26 if(n == 1) printf("1"); 27 else 28 { 29 i = 0; 30 while(n != 1) 31 { 32 if(n % vv[i] == 0) 33 { 34 int tem = 0; 35 while(n % vv[i] == 0) 36 { 37 ++tem; 38 n = n / vv[i]; 39 } 40 41 if(fir) 42 { 43 printf("%d",vv[i]); 44 fir = false; 45 } 46 else 47 { 48 printf("*%d",vv[i]); 49 } 50 51 if(tem > 1) 52 { 53 printf("^%d",tem); 54 } 55 } 56 57 ++i; 58 } 59 } 60 61 printf("\n"); 62 } 63 return 0; 64 }
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原文地址：http://www.cnblogs.com/xiaoyesoso/p/4279571.html
