cf——C. Arithmetic Progression

2024-06-09 04:58

文章标签 cf arithmetic progression

本文主要是介绍cf——C. Arithmetic Progression，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Everybody knows what an arithmetic progression is. Let us remind you just in case that an arithmetic progression is such sequence of numbers a1, a2, ..., an of length n, that the following condition fulfills:

a2 - a1 = a3 - a2 = a4 - a3 = ... = ai + 1 - ai = ... = an - an - 1.

For example, sequences [1, 5], [10], [5, 4, 3] are arithmetic progressions and sequences [1, 3, 2], [1, 2, 4] are not.

Alexander has n cards containing integers. Arthur wants to give Alexander exactly one more card with a number so that he could use the resulting n + 1 cards to make an arithmetic progression (Alexander has to use all of his cards).

Arthur has already bought a card but he hasn't written a number on it. Help him, print all integers that you can write on a card so that the described condition fulfilled.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of cards. The next line contains the sequence of integers — the numbers on Alexander's cards. The numbers are positive integers, each of them doesn't exceed 108.

Output

If Arthur can write infinitely many distinct integers on the card, print on a single line -1.

Otherwise, print on the first line the number of integers that suit you. In the second line, print the numbers in the increasing order. Note that the numbers in the answer can exceed 108 or even be negative (see test samples).

Sample test(s) 
input 
3
4 1 7
output 
2
-2 10
input 
1
10
output 
-1
input 
4
1 3 5 9
output 
1
7
input 
4
4 3 4 5
output 
0
input 
2
2 4
output 
3
0 3 6
分类讨论，怎样成为一个等差数列 
1.若n==0，输出0; 
2.若n==1,输出-1; 
3.若n==2&&a[0]==a[1],输出1，a[0]; 
4.若n==2&&a[0]!=a[1]&&(a[1]-a[0])%2==0,输出3，a[0]-d,a[0]+d/2,a[1]+d; 
5.若n==2&&a[0]!=a[1]&&(a[1]-a[0])%2!=0,输出2，a[0]-d,a[1]+d; 
5.若n>2&&d==0&&count==0,输出1，a[0]; 
6.若n>2&&d!=0&&count=1,(a[flag]-a[flag-1])%2==0，输出1，a[flag-1]+d; 
7.若n>2&&d!=0&&count>1,输出0； 
 #include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int a[100005];
int main()
{int n;while(cin>>n){for(int i=0;i<n;i++)cin>>a[i];sort(a,a+n);if(n==1){cout<<"-1"<<endl;continue;}else if(n==0){cout<<"0"<<endl;continue;}else if(n==2){if(a[0]==a[1])cout<<"1\n"<<a[0]<<endl;else if((a[1]-a[0])%2==0){int d=a[1]-a[0];cout<<"3\n"<<a[0]-d<<" "<<d/2+a[0]<<" "<<a[1]+d<<endl;continue;}else{int d=a[1]-a[0];cout<<"2\n"<<a[0]-d<<" "<<a[1]+d<<endl;continue;}}else if(n>2){int flag=0;int d=9999999999;for(int i=1;i<n;i++){if(a[i]-a[i-1]<d)d=a[i]-a[i-1];}int count=0;for(int i=1;i<n;i++){if(a[i]-a[i-1]!=d){flag=i;count++;}}if(count==0&&d==0)cout<<"1\n"<<a[0]<<endl;else if(count==0&&d!=0)cout<<"2\n"<<a[0]-d<<" "<<a[n-1]+d<<endl;else if(count==1&&a[flag]-a[flag-1]==2*d)cout<<"1\n"<<a[flag-1]+d<<endl;elsecout<<"0\n";}}return 0;
}
 
 
 