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, ..., an (1 ≤ ai ≤ 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.

Examples

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.

题解:去看是否出现次数最多的是否占n个数的一半以上

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream> using namespace std; int main()
{ int n;
cin>>n;
int a[105];
int vis[1005]={0};
memset(vis,0,sizeof(vis));
for(int t=0;t<n;t++)
{
scanf("%d",&a[t]);
vis[a[t]]++;
}
int maxn=0;
for(int t=1;t<=1000;t++)
{
if(vis[t]>=2)
{
if(vis[t]>maxn)
maxn=vis[t];
}
}
if(maxn<=(n+1)/2)
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
return 0;
}

CodeForces - 296A-Yaroslav and Permutations(思维)的更多相关文章

  1. Codeforce 296A - Yaroslav and Permutations

    Yaroslav has an array that consists of n integers. In one second Yaroslav can swap two neighboring a ...

  2. CodeForces - 987E Petr and Permutations (思维+逆序对)

    题意:初始有一个序列[1,2,...N],一次操作可以将任意两个位置的值互换,Petr做3*n次操作:Alxe做7*n+1次操作.给出最后生成的新序列,问是由谁操作得到的. 分析:一个序列的状态可以归 ...

  3. CodeForces - 427A (警察和罪犯 思维题)

    Police Recruits Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Sub ...

  4. Codeforces Round #337 Alphabet Permutations

    E. Alphabet Permutations time limit per test:  1 second memory limit per test:  512 megabytes input: ...

  5. codeforces 341C Iahub and Permutations(组合数dp)

    C. Iahub and Permutations time limit per test 1 second memory limit per test 256 megabytes input sta ...

  6. Codeforces 463D Gargari and Permutations

    http://codeforces.com/problemset/problem/463/D 题意:给出k个排列,问这k个排列的最长公共子序列的长度. 思路:只考虑其中一个的dp:f[i]=max(f ...

  7. codeforces 340E Iahub and Permutations(错排or容斥)

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Iahub and Permutations Iahub is so happy ...

  8. codeforces 895B XK Segments 二分 思维

    codeforces 895B XK Segments 题目大意: 寻找符合要求的\((i,j)\)对,有:\[a_i \le a_j \] 同时存在\(k\),且\(k\)能够被\(x\)整除,\( ...

  9. codeforces 893D Credit Card 贪心 思维

    codeforces 893D Credit Card 题目大意: 有一张信用卡可以使用,每天白天都可以去给卡充钱.到了晚上,进入银行对卡的操作时间,操作有三种: 1.\(a_i>0\) 银行会 ...

  10. C. Nice Garland Codeforces Round #535 (Div. 3) 思维题

    C. Nice Garland time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...

随机推荐

  1. BZOJ 1041 [HAOI2008]圆上的整点:数学【费马平方和定理】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1041 题意: 给定n(n <= 2*10^9),问你在圆x^2 + y^2 = n^ ...

  2. leetcode 290 Word Pattern(map的应用)

    Given a pattern and a string str, find if str follows the same pattern. Here follow means a full mat ...

  3. VC6.0实用小技巧

    VC6.0的若干实用小技巧 .检测程序中的括号是否匹配 把光标移动到需要检测的括号(如大括号{}.方括号[].圆括号()和尖括号<>)前面,键入快捷键 “Ctrl+]”.如果括号匹配正确, ...

  4. Seal Report_20160923

    Seal Report算是报表工具中比较好用的一个,它提供了一个完整的从其他任何数据库产生报表的架构.该产品主要关注于容易安装和报表设计,一旦安装好,报表很快就可以建立并且发布.该组件完全开源,使用C ...

  5. HDU5692 Snacks

    HDU5692 Snacks Problem Description 百度科技园内有n个零食机,零食机之间通过n−1条路相互连通.每个零食机都有一个值v,表示为小度熊提供零食的价值. 由于零食被频繁的 ...

  6. codevs 3314 魔法森林

    传送门 3314 魔法森林  时间限制: 3 s  空间限制: 256000 KB  题目等级 : 大师 Master 题解   题目描述 Description 为了得到书法大家的真传,小E同学下定 ...

  7. 字典树Trie的使用

    1. Trie树介绍 Trie,又称单词查找树.前缀树,是一种多叉树结构.如下图所示: 上图是一棵Trie树,表示了关键字集合{“a”, “to”, “tea”, “ted”, “ten”, “i”, ...

  8. jdk安装图解--windows系统(第一次安装和第二次安装区别)

    第一次安装可参考 https://jingyan.baidu.com/article/22fe7cedc9b93e3003617f64.html 第二次安装,如已经配置好环境变量,cmd下执行java ...

  9. BZOJ1503:[NOI2004]郁闷的出纳员

    浅谈\(splay\):https://www.cnblogs.com/AKMer/p/9979592.html 浅谈\(fhq\)_\(treap\):https://www.cnblogs.com ...

  10. Poj1007_DNA Sorting(面向对象方法)

    一.Description One measure of ``unsortedness'' in a sequence is the number of pairs of entries that a ...