PAT 1041 Be Unique_年龄问题 time limit: 1 sec memory limit: 64 mb submit:-优快云博客

本文链接：https://blog.youkuaiyun.com/ZCMU_2024/article/details/84798143

博客围绕火星独特的彩票规则展开，规则是第一个押注唯一数字者获胜。给出输入输出规范及示例，关键在于找出第一个不重复出现的数字，最后还提及了相关代码。

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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 Nbets. 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

题意:关键句 "The first one who bets on a unique number wins" ,即找出第一个不重复出现的数字

代码

#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
const int Max=1e4+5;
int main()
{
    int Bet[Max],N,x;
    scanf("%d",&N);
    vector<int> Mark;
    for(int i=0;i<N;i++)
    {
        scanf("%d",&x);
        if(!Bet[x])
          Mark.push_back(x);
        Bet[x]+=1;
    }
    for(int i=0;i<Mark.size();i++)
    {
        x=Mark[i];
        if(Bet[x]==1)
        {
            printf("%d",x);
            return 0;
        }
    }
    printf("None");
    return 0;
}