1059 Prime Factors

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

题意：

　　给出一个数，要求写出其质因子乘积的形式，相同的质因子用指数表示。

思路：

　　首先打出一个素数表，然后遍历素数表中的素数，如果该素数是所给数字的因子则记录下来，更新所给数字 temp /= i。如果最后temp > 1，则也要将其输出。

Code：

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int main() {
 6     long long n;
 7     cin >> n;
 8     if (n == 1) {
 9         cout << n << "=" << n;
10         return 0;
11     } else {
12         cout << n << "=";
13     }
14     vector<int> prime(50005, 1);
15     prime[0] = prime[1] = 0;
16     for (int i = 2; i * i < 50000; ++i) {
17         for (int j = 2; i * j < 50000; ++j) {
18             prime[i * j] = 0;
19         }
20     }
21     bool isFirst = true;
22     int temp = n;
23     for (int i = 2; i < 50005; ++i) {
24         int cnt = 0;
25         bool found = false;
26         while (prime[i] == 1 && temp % i == 0) {
27             temp /= i;
28             cnt++;
29             found = true;
30         }
31         if (found) {
32             if (isFirst) {
33                 cout << i;
34                 if (cnt > 1) cout << "^" << cnt;
35                 isFirst = false;
36             } else {
37                 cout << "*" << i;
38                 if (cnt > 1) cout << "^" << cnt;
39             }
40         }
41     }
42     if (temp > 1) cout << (isFirst ? "" : "*") << temp << endl;
43     return 0;
44 }

 

posted @ 2020-05-06 21:02  Veritas_des_Liberty  阅读(182)  评论(0)  编辑  收藏  举报