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:

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=-;

    cin>>n;

    unordered_map<int,int> mp;

    vector<int> vt;

    int key;

    for(int i=;i<n;i++){

        cin>>key;

        vt.push_back(key);

        if(mp[key]==)

            mp[key]=-;

        else

            mp[key]=;

    }

    for(int i=;i<vt.size();i++){

        if(mp[vt[i]]==-){

            ans=vt[i];break;

        }

    }

    if(ans==-)

        cout<<"None";

    else

        cout<<ans;

   return();

}

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。

巴特西