1059. Prime Factors (25)

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

以下算法可以改进为比3大的可以直接加2，忽略偶数

 1 #include <iostream>
 2 #include <vector>
 3
 4 using namespace std;
 5
 6 int main()
 7 {
 8     vector<int> factors, radix;
 9     long factor = 2;
10     long num;
11     cin >> num;
12     cout << num << "=";
13     if (num == 1)
14     {
15         cout << 1;
16         return 0;
17     }
18     while (num != 1)
19     {
20         int cnt = 0;
21         if (num % factor == 0)
22             factors.push_back(factor);
23         while (num%factor == 0)
24         {
25             cnt++;
26             num /= factor;
27         }
28         if (cnt!=0)
29             radix.push_back(cnt);
30         factor++;
31     }
32
33     for (int i = 0; i < factors.size() - 1; i++)
34     {
35         cout << factors[i];
36         if (radix[i]>1)
37             cout << "^" << radix[i];
38         cout << "*";
39     }
40     cout << factors.back();
41     if (radix.back() > 1)
42         cout << "^" << radix.back();
43 }