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 [1,10​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。