博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Codeforces Round #179 (Div. 2) A. Yaroslav and Permutations(简单)
阅读量:5848 次
发布时间:2019-06-18

本文共 1879 字,大约阅读时间需要 6 分钟。

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, ..., 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.

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。

1 #include 
2 #include
3 #include
4 #include
5 #include
6 using namespace std; 7 8 int main() 9 {10 int n, a[101], cnt[1001];11 bool ok;12 while(scanf("%d", &n) != EOF)13 {14 ok = true;15 memset(cnt, 0, sizeof(cnt));16 for(int i = 0; i < n; i++)17 {18 scanf("%d", &a[i]);19 cnt[a[i]]++;20 }21 int max = 0;22 for(int i = 1; i < 1001; i++)23 if(max < cnt[i]) max = cnt[i];24 if((n & 1) && (max > n / 2 + 1)) ok = false;25 else if(!(n & 1) && (max > n / 2)) ok = false;26 if(ok) cout << "YES\n";27 else cout << "NO\n";28 }29 return 0;30 }

 

 

 

 

 

 

转载地址:http://qywjx.baihongyu.com/

你可能感兴趣的文章
介绍2个免费生成手机端软件的网站
查看>>
Objective-C的算术表达式 .
查看>>
RPC failed; result=28, HTTP code = 0
查看>>
gcc编译C++程序
查看>>
linux中nfs的自动挂载
查看>>
统一关闭域客户端防火墙服务/功能
查看>>
expandablelistview open group scroll to top
查看>>
Android通过Aidl调用Service实例
查看>>
找回使用Eclipse删除的文件
查看>>
rabbitmq 消息系统 消息队列
查看>>
Intellij IDEA神器居然还有这些小技巧
查看>>
HBase基本原理
查看>>
php使用qr生成二维码
查看>>
集成spring3、hibernate4、junit
查看>>
eclipse常用快捷键
查看>>
URL与ASCII
查看>>
Jetty9安装部署
查看>>
OpenGL的视图变换
查看>>
Redis.conf 说明
查看>>
shell-用grep查看输入的参数是否在/etc/passwd中
查看>>