PAT A1041

最新推荐文章于 2024-05-31 13:33:10 发布

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博客介绍了火星独特的彩票规则，即第一个押注唯一数字者获胜。给出了时间、内存、代码长度等限制，还说明了输入包含正整数N及N个押注数字，输出为获胜数字，若无则输出“None”。

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1041. Be Unique (20)

时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
#include <vector>
#include <cstring>
#include <cmath>
#include <set>
#include <queue>
#include <algorithm> 
#include <bitset>
#include <sstream> 
#include <iomanip>
#include <map>

using namespace std;


map<int,int> mp;

int n; 



int main(){
    cin>>n;
    int temp[n];

    for(int i=0;i<n;i++){


        cin>>temp[i];
        if( mp.find(temp[i]) == mp.end()){
            mp[temp[i]]=1;

        }else{

            mp[temp[i]]++;
        }

    }

    int res=-1;

    for(int i=0;i<n;i++){

        if(mp[temp[i]] == 1){

            res=temp[i];
            break;
        }


    }

    if(res==-1){
        cout<<"None"<<endl;
    }else{

        cout<<res<<endl;
    }



    return 0;
}