https://codeforces.com/contest/1154/problem/B

题意：给定一个数组，现在有一个操作：对于数组的某个元素，你可以加或者减D（仅一次）。任务是将数组的所有值变成一样的。输出D的值，不存在输出-1。

题解：等差数列+思维

1、特判n==1的时候；

2、特判元素为2个，而且公差为偶数的时候

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N],b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);

    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    b[++cnt]=a[1];
    for(int i=2;i<=n;i++){
        if(a[i-1]!=a[i])b[++cnt]=a[i];
    }
    if(cnt==1){
        cout<<0<<endl;
        return 0;
    }
    int d=(b[2]-b[1]);
    for(int i=3;i<=cnt;i++){
        if(b[i]-b[i-1]!=d){
            cout<<-1<<endl;
            return 0;
        }
    }
    if(cnt==2&&d%2==0){
        d/=2;
    }
    cout<<d<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}