PAT的1041 Be Unique，为什么会 时间超时？！！

题目描述：
PAT 1041 Be Unique
1041 Be Unique
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
提问：

#include <cstdio>
#include <vector>
#include <cmath>
#include <map>
using namespace std;

int main(){

    int n;
    while(scanf("%d",&n)!=EOF){
        vector <int> vi;
        for(int i=0;i<n;++i){
            int  tmp;
            scanf("%d",&tmp);
            if(tmp>=1&&tmp<=pow(10,5))   vi.push_back(tmp);
        }
//        bool bigFlag=false;//判断有没有出现唯一数
        int i;
        for(i=0;i<n-1;++i){
            bool flag=false;    //vi[i]是唯一
            for(int j=i+1;j<n;j++){
                if(vi[i]==vi[j]) {
                    vi[j]=0;
                    flag=true;
                }
            }
            if(flag)    vi[i]=0; //不是唯一
            else{//是唯一
                printf("%d\n",vi[i]);
//                bigFlag=true;
                break;
            }
        }//for
        if(i==n-1) printf("None\n");  //整个数组没有唯一的
        vi.clear();

    }
    return 0;
}

https://blog.csdn.net/linghugoolge/article/details/83280698

时间复杂度高了

这题用散列完全可以O(n)解决