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

---

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10^?4]. The first one who bets on a unique number wins. For example, if there are 7 people betting on { 5 31 5 88 67 88 17 }, then the second one who bets on 31 wins.

Input Specification:
Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10^?5) and then followed by N bets. The numbers are separated by a space.

Output Specification:
For each test case, print the winning number in a line. If there is no winner, print None instead.

Sample Input 1:
7 5 31 5 88 67 88 17
Sample Output 1:
31

Sample Input 2:
5 888 666 666 888 888
Sample Output 2:
None

---

題意：給一串數，輸出第一個只出現過一次的數，若無則輸出None

分析：定義隊列存儲數據，定義map數組查重。輸出第一個不重復的即可。 若無則輸出None

---

#include<bits/stdc++.h> using namespace std; int main( ) {int n; cin >> n;queue<int>q;map<int, int>m;while(n--) {int x; cin >> x; q.push(x); m[x]++;}while(!q.empty()) {if(m[q.front()] == 1) { cout << q.front() << '\n'; goto loop; }q.pop(); }cout << "None\n";loop :;return 0; }

---

耗時：

---

求贊哦~

總結

以上是生活随笔為你收集整理的【题意+解析】1041 Be Unique (20 分)_18行代码AC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。