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

 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 }