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
- Hibernate5.2之原生SQL查询
Hibernate5.2之原生SQL查询 一. 介绍 在上一篇博客中笔者通过代码的形式给各位读者介绍了Hibernate中最重要的检索方式--HQL查询.在本博文中笔者将向各位读者介绍Hiberna ...
- PLSQL 逻辑多线程
PROCEDURE get_sheetid(i_topic IN VARCHAR2, o_newsheetid OUT VARCHAR2) IS PRAGMA AUTONOMOUS_TRANSA ...
- Java中的枚举
枚举通过enum关键字进行定义.实际上当定义一个枚举类型的时候,该类型默认继承了Enum类. 枚举的定义格式如下: enum Color { RED,BLUE,GREEN; } 下面演示一个枚举变量的 ...
- Oracle 收缩表大小 Oracle Shrink Table --转载
从10g开始,oracle开始提供Shrink的命令,假如我们的表空间中支持自动段空间管理 (ASSM),就可以使用这个特性缩小段,即降低HWM.这里需要强调一点,10g的这个新特性,仅对ASSM表空 ...
- weblogic 优化设置 http://wenku.baidu.com/view/c42e7a5bbe23482fb4da4cf2.html
引自:http://wenku.baidu.com/view/c42e7a5bbe23482fb4da4cf2.html
- c#版在pc端发起微信扫码支付
等了好久,微信官方终于发布了.net的demo. 主要代码: /** * 生成直接支付url,支付url有效期为2小时,模式二 * @param productId 商品ID * @return 模式 ...
- MySQL绿色版5.7以上安装教程
写在前面:5.7增加了安全性,默认root密码不在为空,而是初始化时随机生成一个root密码,改root密码的方式也不一样了 下载地址 http://dev.mysql.com/downloads/m ...
- location.href跳转不正确
今天写这个随笔的用意是为了记录我遇到的一种情况,导致我页面无法正确跳转 location.href跳转页面其实很简单,只要附上url就可以了,但是今天我在测试一个跳转时是这么写的: location. ...
- boost compile
pushd E:\boost\boost_1_59_0 b2 stage --toolset=msvc-12.0 --without-python --stagedir="E:\boost\ ...
- Analyze network packet files very carefully
As a professional forensic guy, you can not be too careful to anlyze the evidence. Especially when t ...