PAT 1041 Be Unique[简单]
1041 Be Unique (20 分)
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 [1,104]. 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 (≤105) 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
题目大意:给出n个数,判断其中不重复出现的第一个数,如果均是重复出现,那么就输出None.
//还是比较简单的。AC了:
#include <iostream> #include <vector> #include<unordered_map> using namespace std; int main() { int n,ans=-1; cin>>n; unordered_map<int,int> mp; vector<int> vt; int key; for(int i=0;i<n;i++){ cin>>key; vt.push_back(key); if(mp[key]==0) mp[key]=-1; else mp[key]=1; } for(int i=0;i<vt.size();i++){ if(mp[vt[i]]==-1){ ans=vt[i];break; } } if(ans==-1) cout<<"None"; else cout<<ans; return(0); }
1.其实可以不使用unorder_map的,它并不是按输入顺序排序,而是随机的吧,可以使用map
2.既然要记录顺序,那么就使用vector来存储原来的输入顺序这个是需要的。