Codeforces 723C. Polycarp at the Radio 模拟
2 seconds
256 megabytes
standard input
standard output
Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a1, a2, ..., an, where ai is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to m, but he doesn't really like others.
We define as bj the number of songs the group j is going to perform tomorrow. Polycarp wants to change the playlist in such a way that the minimum among the numbers b1, b2, ..., bm will be as large as possible.
Find this maximum possible value of the minimum among the bj (1 ≤ j ≤ m), and the minimum number of changes in the playlist Polycarp needs to make to achieve it. One change in the playlist is a replacement of the performer of the i-th song with any other group.
The first line of the input contains two integers n and m (1 ≤ m ≤ n ≤ 2000).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the performer of the i-th song.
In the first line print two integers: the maximum possible value of the minimum among the bj (1 ≤ j ≤ m), where bj is the number of songs in the changed playlist performed by the j-th band, and the minimum number of changes in the playlist Polycarp needs to make.
In the second line print the changed playlist.
If there are multiple answers, print any of them.
4 2
1 2 3 2
2 1
1 2 1 2
7 3
1 3 2 2 2 2 1
2 1
1 3 3 2 2 2 1
4 4
1000000000 100 7 1000000000
1 4
1 2 3 4
In the first sample, after Polycarp's changes the first band performs two songs (b1 = 2), and the second band also performs two songs (b2 = 2). Thus, the minimum of these values equals to 2. It is impossible to achieve a higher minimum value by any changes in the playlist.
In the second sample, after Polycarp's changes the first band performs two songs (b1 = 2), the second band performs three songs (b2 = 3), and the third band also performs two songs (b3 = 2). Thus, the best minimum value is 2.
题目连接:http://codeforces.com/contest/723/problem/C
题意:改变播放清单里面的尽量少的播放歌曲,使得1~m播放次数最小的尽可能大。
思路:播放清单里面,1~m里面每首歌至少有播放n/m遍。
#include<bits/stdc++.h>
using namespace std;
int a[];
int sign[],flag[];
int main()
{
int i,j,n,m;
scanf("%d%d",&n,&m);
for(i=; i<=n; i++)
{
scanf("%d",&a[i]);
if(a[i]<=m) sign[a[i]]++;
}
int cou=n/m,amount=n%m;
int ans=;
for(i=; i<=n; i++)
{
if(a[i]<=m&&sign[a[i]]<cou+) continue;
if(a[i]<=m&&sign[a[i]]==cou&&amount>=)
{
if(flag[a[i]]==) amount--;
flag[a[i]]=;
continue;
}
for(j=; j<=m; j++)
if(sign[j]<cou)
{
if(a[i]<=m) sign[a[i]]--;
sign[j]++;
a[i]=j;
ans++;
break;
}
}
cout<<cou<<" "<<ans<<endl;
for(i=; i<=n; i++)
cout<<a[i]<<" ";
cout<<endl;
return ;
}
Codeforces 723C. Polycarp at the Radio 模拟的更多相关文章
- codeforces 723C : Polycarp at the Radio
Description Polycarp is a music editor at the radio station. He received a playlist for tomorrow, th ...
- CodeForces 723C Polycarp at the Radio (题意题+暴力)
题意:给定 n 个数,让把某一些变成 1-m之间的数,要改变最少,使得1-m中每个数中出现次数最少的尽量大. 析:这个题差不多读了一个小时吧,实在看不懂什么意思,其实并不难,直接暴力就好,n m不大. ...
- Codeforces Round #375 (Div. 2) C. Polycarp at the Radio 贪心
C. Polycarp at the Radio time limit per test 2 seconds memory limit per test 256 megabytes input sta ...
- Codeforces 659F Polycarp and Hay 并查集
链接 Codeforces 659F Polycarp and Hay 题意 一个矩阵,减小一些数字的大小使得构成一个连通块的和恰好等于k,要求连通块中至少保持一个不变 思路 将数值从小到大排序,按顺 ...
- 【23.48%】【codeforces 723C】Polycarp at the Radio
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【Codeforces 723C】Polycarp at the Radio 贪心
n个数,用最少的次数来改变数字,使得1到m出现的次数的最小值最大.输出最小值和改变次数以及改变后的数组. 最小值最大一定是n/m,然后把可以改变的位置上的数变为需要的数. http://codefor ...
- Codeforces Round #375 (Div. 2) Polycarp at the Radio 优先队列模拟题 + 贪心
http://codeforces.com/contest/723/problem/C 题目是给出一个序列 a[i]表示第i个歌曲是第a[i]个人演唱,现在选出前m个人,记b[j]表示第j个人演唱歌曲 ...
- Codeforces Round #528-A. Right-Left Cipher(字符串模拟)
time limit per test 1 second memory limit per test 256 megabytes input standard input output standar ...
- cf723c Polycarp at the Radio
Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be re ...
随机推荐
- SpringMVC学习系列(5) 之 数据绑定-2
在系列(4)中我们介绍了如何用@RequestParam来绑定数据,下面我们来看一下其它几个数据绑定注解的使用方法. 1.@PathVariable 用来绑定URL模板变量值,这个我们已经在系列(3) ...
- IIS-Server is too busy _解决方法
httpRuntime Server Too Busy 修改方法:修改服务器.net配置“machine.config"文件,该文件位于Windows系统目录下,如“C:\WINDOWS \Micro ...
- WeX5之xid相关API
WeX5针对xid提供了以下js api: 1.根据xid获取id:this.getIDByXID(xid): 2.根据xid获取HTML节点:this.getElementByXid(xid),此a ...
- 每天一个 Linux 命令(4):mkdir
linux mkdir 命令用来创建指定的名称的目录,要求创建目录的用户在当前目录中具有写权限,并且指定的目录名不能是当前目录中已有的目录. 1.命令格式: mkdir [选项] 目录- 2.命令功能 ...
- 制作HP MicroServer Gen8可用的ESXi 5.x SD/TF卡启动盘
前些日子看到HP公司和京东在搞服务器促销活动,于是就入了一个 ProLiant MicroServer Gen8 的低配版 相比上一代产品,新一代 MicroServer系列微型服务器可更换处理器,还 ...
- Mono addin 学习笔记 3
典型的基于Mono addin插件框架的应用程序有以下一个部分组成: 1. 主应用程序:提供了一系列的扩展点(Extension Point)供其他应用进行扩展: 2. 扩展插件: 其部署结构图如下为 ...
- GLSL语言基础
from http://www.kankanews.com/ICkengine/archives/120870.shtml 变量 GLSL的变量命名方式与C语言类似.变量的名称可以使用字母,数字以及下 ...
- linux下的磁盘和文件系统管理
一.硬盘分区知识 1.分区类型 硬盘分区一共有3种:主分区.扩展分区和逻辑分区.扩展分区只不过是逻辑分区的“容器”,实际上只有主分区和逻辑分区进行数据存储.在一块硬盘上最多只能有4个主分区,可以另外建 ...
- VBS基本知识
由于一些需要,开始学习VBS了.此篇文章一直将处于编辑添加状态. 1.VBS简介 VBS 即VBScript(Microsoft Visual Basic Script Editon),是微软开发的一 ...
- Selenium + PhantomJS + python 简单实现爬虫的功能
Selenium 一.简介 selenium是一个用于Web应用自动化程序测试的工具,测试直接运行在浏览器中,就像真正的用户在操作一样 selenium2支持通过驱动真实浏览器(FirfoxDrive ...