Codeforces Round #179 (Div. 2) A. Yaroslav and Permutations（简单）

A. Yaroslav and Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Yaroslav has an array that consists of n integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.

Help Yaroslav.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array. The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000) — the array elements.

Output

In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.

Sample test(s)

Input

1
1

Output

YES

Input

3
1 1 2

Output

YES

Input

4
7 7 7 7

Output

NO

Note

In the first sample the initial array fits well.

In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.

In the third sample Yarosav can't get the array he needs.

 这是一道很简单的题。。。不过注意题目的理解：元素是可以交换多次的。分n为奇数和偶数两种情况讨论即可。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

int main()
{
    int n, a[101], cnt[1001];
    bool ok;
    while(scanf("%d", &n) != EOF)
    {
        ok = true;
        memset(cnt, 0, sizeof(cnt));
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            cnt[a[i]]++;
        }
        int max = 0;
        for(int i = 1; i < 1001; i++)
            if(max < cnt[i]) max = cnt[i];
        if((n & 1) && (max > n / 2 + 1)) ok = false;
        else if(!(n & 1) && (max > n / 2)) ok = false;
        if(ok) cout << "YES\n";
        else cout << "NO\n";
    }
    return 0;
}