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要記得考慮找不到公差的情況，就代表缺少的數字一定在中間(如我只有五項的話)

解法:
#include <iostream>

#include <vector>

#include <algorithm>

#include <string>

#include <sstream>

using namespace std;

 

int main() {

int t;

cin >> t;

string s;

getline(cin ,s);

vector<int>v;

while(t--){

getline(cin ,s);

int n,m,d,dd;

istringstream iss(s);

while(iss >> n){

v.push_back(n);

}

sort(v.begin(),v.end());

bool found=false;

for(int i=0;i<v.size();i++){

if(v[i]==v[i+1]){

d=v[i];

v[i]=v[i-1];

}

else if((v[i-1]+v[i+1])/2 == v[i]){

m = v[i]-v[i-1];//公差

for(int i=0;i<v.size();i++){

if(v[i+1] - v[i] == m*2){

m=v[i+1]-m;

}

}

found=true;

}

}

if(!found){

int g=(v[0]+v[v.size()-1])/2;

cout << g <<' ';

}

else{

cout << m << ' ';

}

cout <<d <<'\n';

v.clear();

}

}