CodeForces - 296A-Yaroslav and Permutations(思维)
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
- Codeforce 296A - Yaroslav and Permutations
Yaroslav has an array that consists of n integers. In one second Yaroslav can swap two neighboring a ...
- 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 ...
- CodeForces - 427A (警察和罪犯 思维题)
Police Recruits Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Sub ...
- Codeforces Round #337 Alphabet Permutations
E. Alphabet Permutations time limit per test: 1 second memory limit per test: 512 megabytes input: ...
- 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 ...
- 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 ...
- 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 ...
- 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,\( ...
- 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 ...
- 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
- css3线性渐变兼容
火狐浏览器: background:-moz-linear-gradient(top, red, rgba(0, 0, 255, 0.5)); 谷歌: .l6{background: -webkit- ...
- laravel基础课程---15、分页及验证码(lavarel分页效果如何实现)
laravel基础课程---15.分页及验证码(lavarel分页效果如何实现) 一.总结 一句话总结: 数据库的paginate方法:$data=\DB::table("user" ...
- 高并发压力下导致数据库bug
环境信息: linux 6.1 + oracle11.2.0.3 RAC 问题现象: 学校晚上6点选课,人数大概有3000,7点时,数据库报错如下(数据库到6点多还是可以连接的),数据库hu ...
- 初识Spacy
之所以想接触Spacy,是看到其自称为工业级的应用,所以想尝试下 windows下安装Spacy: 直接安装pip install spacy是会报错的 解决方法: 到 htt ...
- PC样式reset参考
/* html5doctor.com Reset Stylesheet */ * { padding:; margin:; list-style: none; } html, body, div, s ...
- 每天一个linux命令(5):mkdir命令
版权声明更新:2017-05-09博主:LuckyAlan联系:liuwenvip163@163.com声明:吃水不忘挖井人,转载请注明出处! 1 文章介绍 本文介绍了Linux下命令mkdir. 2 ...
- POCO库中文编程参考指南(10)如何使用TCPServer框架?
1 TCPServer 框架概述 POCO 库提供TCPServer框架,用以搭建自定义的 TCP 服务器.TCPServer维护一个连接队列.一个连接线程池.连接线程用于处理连接,连接线程只要一空闲 ...
- Lagom学习 (二)
以一个官方的例子,开启lagom的学习之旅. 1: git clone https://github.com/lagom/activator-lagom-java-chirper.git. 2: ...
- C#API接口调试工具
自从去年软件界网站开发推崇前后端分离,我们公司也在进行转行行,从原先的前端架构,后端架构,数据库搭建一肩挑的模式也逐渐转型为前后端分离,大量招收前端开发人员,原来的人员也转型为专职的后端开发,这样的变 ...
- Java异常控制机制和异常处理原则【转】
原文:https://www.jianshu.com/p/15872cba211d Java异常控制机制又被称为“违例控制机制”. 捕获程序错误最理想的时机是在编译阶段,这样可以彻底避免错误的代码运行 ...