求出最大最小值，从小到大遍历，将出现的次数计入a【n】，最后输出重复和缺失的数字
···

#include<iostream>
#include<sstream>
#include<cstring>
#include <algorithm>
using namespace std;
const int N = 100100;
int n;
int a[N];

   int main(){
     int c,d,min1= 10010,max1=0;
     int n,m;
     cin >> c;
     while(c--){
         while(cin >> d){
             a[d]++;
             min1 = min(min1,d);
             max1 = max(max1,d);
         }
     }
     for(int i = min1;i < max1;i ++){
         if(a[i] == 0) m = i;
         else if(a[i] == 2) n = i;
     }
     cout << m <<" " << n;
    return 0;
}

```