1041 Be Unique（20 分）

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原创最新推荐文章于 2025-12-10 16:41:55 发布 464 阅读 AI 写同款· GEO 优化 ›

单片机按键识别篇---单击---双击----长按　　最近做一任务需要使用一个按键实现三种功能，分别是按键单击功能，按键双击功能和按键长按功能，可能之前没有接触过这类按键复用情况，顶多也只是简单识别单击和长按，没有想过双击以上按键功能，也是绞尽脑汁，想了半天，又上网看了些例程，算是对于按键的识别有点小小的了解，感觉这个功能挺实用，在此做个随笔。一、思路　　1.识别长按操作思路：我们使用系统定时器定时，然后让系统每过一段时间去扫描按键I/O... 阅读详情

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 [1,10
4
]. 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 (≤10
5
) 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

解题思路：
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。

#include<cstdio>
const int maxn=1e5+10;
int book[maxn]={0} ;
int a[maxn];
int main(){
    int n,x;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&x);
        a[i]=x;
        book[x]++;
    }
    int ans=-1;
    for(int i=0;i<n;i++){
        if(book[a[i]]==1){
            ans=a[i];
            break;
        }
    }
    if(ans==-1) printf("None\n");
    else printf("%d\n",ans);
    return 0;
}