CUIT ACM Personal Training 11.27(FM）D - Yaroslav and Permutations

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2016年11月28日CUIT ACM 训练赛题解专栏收录该内容

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探讨如何通过交换操作使数组中所有相邻元素变得互不相同，给出具体实现算法及示例。

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D - Yaroslav and Permutations

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

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 Input

Input

1
1

Output

YES

Input

3
1 1 2

Output

YES

Input

4
7 7 7 7

Output

NO

Hint

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.

题解：此题的题意是给你一串数字，让你判断能否通过交换，使得他们相邻彼此不相同。

那么这个其实就是一个模拟插空的过程，首先挑出出现次数最多的数字，然后将其排成一排，用其他数字一个一个去插空，那么这些数字就有可能符合题意。

情况一：6 5 4 5 3 5出现的次数最多，数量2<=(5+1)/2,所以成立；

情况二：5 4 5 4 5 5出现的次数最多，数量3<=(5+1)/2,所以成立；

情况二：5 4 5 5 5 5出现的次数最多，数量4>(5+1)/2,所以不成立；

所以我们可以很清楚地看出来(n+1)/2是判断点。

AC代码如下:

#include <iostream>
#include <cstdio>
using namespace std;

int n,A[1001],temp;
int maxnum;

int main()
{
	scanf("%d",&n);
	maxnum=0;
	for(int i=0;i<n;i++){
		scanf("%d",&temp); A[temp]++;
		if(A[temp]>maxnum) maxnum=A[temp];
	}
	if(maxnum>(n+1)/2) printf("NO\n");
	else printf("YES\n");
	return 0;
}