1041. Be Unique (20)

叁歲伎倆 2022-05-31 10:49 102阅读 0赞

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

--------------------

题目大意：

对火星上的人来说，与众不同是非常重要的。甚至他们的彩票也是以一种独特的方式设计的。获胜的规则很简单：押注于从\[1,10的4次方\]中选择数字。第一个投注唯一数字的人获胜。例如，如果有7个人投注5 31 5 88 67 88 17, 那么第二个投注32的人获胜。  
输入规格：  
每个输入文件包含一个测试用例。每个测试用例都是以一个正整数开头（<=10的5次方），然后是N个投注。数字被一个空格隔开。  
输出规范：  
对于每个测试用例，在一行中打印获胜的数字。如果没有赢家，则打印“None”。

代码：

#include<stdio.h>
    int a[100001],b[100001];
    int main()
    {
        int i,j,n,m,k,t;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
            a[b[i]]++;
        }
        for(i=0;i<n;i++)
        {
            if(a[b[i]]==1)
            {
                printf("%d\n",b[i]);
                break;
            }
        }
        if(i>=n)
            printf("None\n");
        return 0;
    }