立志用最少的代碼做最高效的表達

---

PAT甲級最優題解——>傳送門

---

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 Nin 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=2^211171011291

---

題意：將一個數拆分成若干質因子相乘，升序輸出。

時間復雜度計算：一百毫秒約運行30萬次數據。 若時間復雜度為O(nlogn)，在最壞條件下，也就是取int上限值21e，峰值運算次數也不會超過10萬。因此不會超時。

具體算法邏輯請閱讀代碼體會。不明白的請評論區留言

---

#include<bits/stdc++.h> using namespace std; using gg = long long; int main() {gg n, n1; cin >> n;n1 = n;cout << n << "=";bool flag = false;for(gg i = 2; i*i <= n1; i++) {if(n%i == 0) {gg k = 0;if(!flag) flag = true;else cout << "*";while(n%i == 0) {k++;n /= i; }cout << i;if(k != 1) cout << '^' << k;}}if(!flag) cout << n1; //當沒有因子時輸出其自身 return 0; }

---

耗時：

---

求贊哦~ (?ω?)

超強干貨來襲 云風專訪：近40年碼齡，通宵達旦的技術人生

總結

以上是生活随笔為你收集整理的【题目分析】1059 Prime Factors (25 分)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。