立志用最少的代碼做最高效的表達(dá)

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PAT甲級最優(yōu)題解——>傳送門

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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 = 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

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題意：將一個數(shù)拆分成若干質(zhì)因子相乘，升序輸出。

時間復(fù)雜度計(jì)算：一百毫秒約運(yùn)行30萬次數(shù)據(jù)。 若時間復(fù)雜度為O(nlogn)，在最壞條件下，也就是取int上限值21e，峰值運(yùn)算次數(shù)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?jiān)u論區(qū)留言

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#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(dāng)沒有因子時輸出其自身 return 0; }

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耗時：

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求贊哦~ (?ω?)

超強(qiáng)干貨來襲 云風(fēng)專訪：近40年碼齡，通宵達(dá)旦的技術(shù)人生

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