日韩性视频-久久久蜜桃-www中文字幕-在线中文字幕av-亚洲欧美一区二区三区四区-撸久久-香蕉视频一区-久久无码精品丰满人妻-国产高潮av-激情福利社-日韩av网址大全-国产精品久久999-日本五十路在线-性欧美在线-久久99精品波多结衣一区-男女午夜免费视频-黑人极品ⅴideos精品欧美棵-人人妻人人澡人人爽精品欧美一区-日韩一区在线看-欧美a级在线免费观看

歡迎訪問 生活随笔!

生活随笔

當(dāng)前位置: 首頁 > 编程资源 > 编程问答 >内容正文

编程问答

【题意+解析】1041 Be Unique (20 分)_18行代码AC

發(fā)布時間:2024/2/28 编程问答 28 豆豆
生活随笔 收集整理的這篇文章主要介紹了 【题意+解析】1041 Be Unique (20 分)_18行代码AC 小編覺得挺不錯的,現(xiàn)在分享給大家,幫大家做個參考.

立志用最少的代碼做最高效的表達(dá)


PAT甲級最優(yōu)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?dāng)?shù),輸出第一個只出現(xiàn)過一次的數(shù),若無則輸出None

分析:定義隊列存儲數(shù)據(jù),定義map數(shù)組查重。輸出第一個不重復(fù)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(jié)

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(nèi)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(wǎng)站內(nèi)容還不錯,歡迎將生活随笔推薦給好友。