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:

31

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[10005]; 
int shu[100005];
int main()
{
    int n;
    cin>>n;
    memset(a,0,sizeof(a));
    for(int i=1;i<=n;i++){
        cin>>shu[i];
        a[shu[i]]++;
    }
    int f=0;
    for(int i=1;i<=n;i++){
        if(a[shu[i]]==1){
            cout<<shu[i];
            f=1;
            break;
        }
    }
    if(!f) cout<<"None";
    return 0;
}