1041 Be Unique (20 分) hash输出第一个出现只出现一次的数字

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

题意：输入n个数，输出第一个出现只出现一次的数字，如果没有这些数字，则输出None。
思路：先用vector存储输入的序列，然后再用hash数组来存储这些数字出现的次数，每次都从0循环到n-1即可。

#include<iostream>
#include<cstdio>
#include<stack> 
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>using namespace std;int n;
int hash1[10010];int main() {
    scanf("%d",&n);vector<int> v(n);for(int i = 0;i < n;i++) {
    scanf("%d",&v[i]);}for(int i = 0;i < n;i++) {
    hash1[v[i]]++;}for(int i = 0;i < n;i++) {
    if(hash1[v[i]] == 1) {
    printf("%d\n",v[i]);return 0;}}printf("None\n");return 0;
}