1041 Be Unique(20分)--PAT甲级 2019-10-10 拿起笔来 暂无评论 1889 次阅读 本文发布于2019-10-10， 文章内容或资源可能已经失效，仅供参考，谢谢。 <h3>Be Unique</h3> <b>题目</b> <blockquote>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<sup>4</sup>]. <!--more-->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.</blockquote> - 时间限制: 400 ms - 内存限制: 64 MB - 代码长度限制: 16 KB **Input Specification:** <pre> Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10<sup>5</sup>) and then followed by N bets. The numbers are separated by a space. </pre> **Output Specification:** <pre> For each test case, print the winning number in a line. If there is no winner, print 'None' instead. </pre> > **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 ``` **我的作答** 编译器：C++(g++) ```cpp #include <iostream> using namespace std; int main() { int N, i; short number[100000], tag[10000]; cin >> N; for(i = 0; i < N; ++i) { cin >> number[i]; switch(tag[number[i]]) { case -1: break; case 0: tag[number[i]] = 1; break; case 1: tag[number[i]] = -1; break; } } for(i = 0; i < N; ++i) { if(tag[number[i]] == 1) { cout << number[i]; return 0; } } cout << "None"; return 0; } ``` 标签: C/C++, PAT 如果您对此页面有任何问题或建议，请在本站留言，或联系邮箱me[at]zkk.me本网站部分内容转载自其他网站，如有侵权，请联系博主