PAT 甲级 1041 Be Unique (20 分)（简单,一遍过）

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]. 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 (≤) 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

题意：

给出n个数字，要求输出第一个只出现一次的数字，如果不存在，输出None.

题解：

用类似打表的方法，统计每个数出现的次数。按输出顺序遍历，当有次数为1的数时，输出，没有就输出None。

AC代码：

#include<iostream>

#include<algorithm>

#include<vector>

#include<queue>

#include<map>

#include<string>

#include<cstring>

using namespace std;

int a[];

int shu[];

int main()

{

    int n;

    cin>>n;

    memset(a,,sizeof(a));

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

        cin>>shu[i];

        a[shu[i]]++;

    }

    int f=;

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

        if(a[shu[i]]==){

            cout<<shu[i];

            f=;

            break;

        }

    }

    if(!f) cout<<"None";

    return ;

}

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