【题意+解析】1041 Be Unique (20 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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

题意：给一串数，输出第一个只出现过一次的数，若无则输出None

分析：定义队列存储数据，定义map数组查重。输出第一个不重复的即可。若无则输出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;
}

耗时：

求赞哦~

【题意+解析】1041 Be Unique (20 分)_18行代码AC相关推荐

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【题意+解析】1041 Be Unique (20 分)_18行代码AC

立志用最少的代码做最高效的表达

耗时：

求赞哦~

【题意+解析】1041 Be Unique (20 分)_18行代码AC相关推荐

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