PAT A1041

• 题目：
  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

• 分析
  题目大意： 给出一系列1~10000的整数，从中找到第一个出现一次的数。若可以找到这样的数，则输出此数；否则，则输出None;

解题思路：
1.数组map1存放每个输出数字的出现次数，初始值为0；
2.query数组按照输入顺序存放数字，并作为map1的下标，可以让map1按照输入顺序查看每个数字出现的次数

代码实现：

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#include <cstdio>
int Hash[10010] = {0}, A[100001];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &A[i]);
        Hash[A[i]]++;
    }
    for(int i = 0; i < n; i++){
        if(Hash[A[i]] == 1){
           printf("%d", A[i]);
           return 0;
        }
    }
    printf("None");
    return 0;
}