Codeforces 371B - Filya and Homework(思维)

B. Filya and Homework

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.

Filya is given an array of non-negative integers a1, a2, …, an. First, he pick an integer x and then he adds x to some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.

Now he wonders if it’s possible to pick such integer x and change some elements of the array using this x in order to make all elements equal.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the number of integers in the Filya’s array. The second line contains n integers a1, a2, …, an (0 ≤ ai ≤ 109) — elements of the array.

Output

If it’s impossible to make all elements of the array equal using the process given in the problem statement, then print “NO” (without quotes) in the only line of the output. Otherwise print “YES” (without quotes).

Examples

input

5

1 3 3 2 1

output

YES

input

5

1 2 3 4 5

output

NO

Note

In the first sample Filya should select x = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.

题意:

给出n个数,对于这n个数,每个都可以选择加K,减K,或者不操作.

判断最终能否使得他们都相等.

解题思路:

先对所有的进行排序,然后去重,如果剩下的元素少于2,那么是可行的.

如果剩下三个,并且min+max == 2*mid,那也是可行的.

这里的unique函数很好用,该算法删除相邻的重复元素，然后重新排列输入范围内的元素，并且返回一个迭代器（容器的长度没变，只是元素顺序改变了），表示无重复的值范围得结束。

AC代码:

#include<stdio.h>
#include<algorithm>
using namespace std;
const int maxn = 100005;
int qdu[maxn];
int main()
{
int n;
scanf("%d",&n);
for(int i = 0;i < n;i++)    scanf("%d",&qdu[i]);
sort(qdu,qdu+n);
int cnt = unique(qdu,qdu+n)-qdu;
if(cnt <= 2)
{
printf("YES\n");
return 0;
}
else if(cnt == 3 && qdu[0] + qdu[2] == qdu[1] << 1)
{
printf("YES\n");
return 0;
}
printf("NO\n");
return 0;
}