time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a sequence a1,a2,…,ana1,a2,…,an consisting of nn integers.
You can choose any non-negative integer DD (i.e. D≥0D≥0), and for each aiai you can:
- add DD (only once), i. e. perform ai:=ai+Dai:=ai+D, or
- subtract DD (only once), i. e. perform ai:=ai−Dai:=ai−D, or
- leave the value of aiai unchanged.
It is possible that after an operation the value aiai becomes negative.
Your goal is to choose such minimum non-negative integer DD and perform changes in such a way, that all aiai are equal (i.e. a1=a2=⋯=ana1=a2=⋯=an).
Print the required DD or, if it is impossible to choose such value DD, print -1.
For example, for array [2,8][2,8] the value D=3D=3 is minimum possible because you can obtain the array [5,5][5,5] if you will add DD to 22 and subtract DD from 88. And for array [1,4,7,7][1,4,7,7] the value D=3D=3 is also minimum possible. You can add it to 11 and subtract it from 77 and obtain the array [4,4,4,4][4,4,4,4].
Input
The first line of the input contains one integer nn (1≤n≤1001≤n≤100) — the number of elements in aa.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100) — the sequence aa.
Output
Print one integer — the minimum non-negative integer value DD such that if you add this value to some aiai, subtract this value from some aiai and leave some aiai without changes, all obtained values become equal.
If it is impossible to choose such value DD, print -1.
Examples
input
Copy
6 1 4 4 7 4 1
output
Copy
3
input
Copy
5 2 2 5 2 5
output
Copy
3
input
Copy
4 1 3 3 7
output
Copy
-1
input
Copy
2 2 8
output
Copy
3
因为数据量就100所以就一个一个枚举就行了
下面是我写的，不知道是什么原理竟然过了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
set<int> s;
int a[100];
int main()
{
int n;
scanf("%d",&n);
if (n==1)
{
printf("0\n");
return 0;
}
for (int i=0;i<n;i++)
{
int x;
scanf("%d",&x);
s.insert(x);
}
int len = s.size();
int k = 0;
set<int>:: iterator it;
for (it=s.begin();it!=s.end();it++)
{
a[k++] = *it;
}
/*for (int i=0;i<len;i++)
{
//if (a[i]==z) cnt++;
cout<<a[i]<<" ";
}
puts("");*/
int c = a[len-1] - a[0];
if (c%2==0) c /= 2;
int z = a[len-1] - c;
for (int i=0;i<len;i++)
{
if (a[i]<z) a[i] += c;
else if (a[i]>z) a[i] -= c;
}
int cnt = 0;
for (int i=0;i<len;i++)
{
if (a[i]==z) cnt++;
}
if (cnt==len) printf("%d\n",c);
else printf("-1\n");
return 0;
} 下面是暴力枚举做的
#include<bits/stdc++.h>
using namespace std;
int a[110];
int main()
{
int n;
scanf("%d",&n);
for (int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int flg = 0;
for (int i=0;i<=100;i++)
{
int f = 1;
int temp = a[0] + i;
for (int j=1;j<n;j++)
{
if (a[j]==temp || a[j]+i==temp || a[j]-i==temp)
continue;
else
{
f = 0;
break;
}
}
if (f==1)
{
flg = 1;
printf("%d\n",i);
break;
}
}
if (flg==0) printf("-1\n");
return 0;
}