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
- remote_addr和x_forwarded_for
做网站时经常会用到remote_addr和x_forwarded_for这两个头信息来获取客户端的IP,然而当有反向代理或者CDN的情况下,这两个值就不够准确了,需要调整一些配置. 什么是remote ...
- oracle11g安装和基本的使用-转载
一.测试操作系统和硬件环境是否符合,我使用的是win2008企业版.下面的都是step by step看图就ok了,不再详细解释. 请留意下面的总的设置步骤:--------------------- ...
- python xlwt,xlutils 在excel里面如何插入一行数据
就是把插入行之后值重新输出来. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 import xl ...
- ios7 indexPathForCell 的坑(真是一个大大的坑)
笔者在编写APP 有一个功能点击cell上一个button,修改cell的在tableview中的位置 在ios8上没有问题. 在ios7上总是崩溃 以下是崩溃后提示: Terminating app ...
- READONLY、、cursor、、VARYING
针对 Transact-SQL 过程的准则:所有 Transact-SQL 数据类型都可以用作参数.您可以使用用户定义的表类型创建表值参数.表值参数只能是 INPUT 参数,并且这些参数必须带有 RE ...
- Chap5:32– 34
32. 从 1 到 n 整数中 k (0,1, 2, 3, 4, 5, 6, 7, 8, 9)出现的次数. 时间 O(log10N) A. 当 K != 0 时: 以 n = 2014,K = 1 ...
- 剑指Offer:面试题21——包含min函数的栈(java实现)
问题描述: 定义栈的数据结构,请在该类型中实现一个能够得到栈的最小元素的min函数.在该栈中,调用min,push及pop的时间复杂度都是O(1). 思路:加入一个辅助栈用来存储最小值集合 (这里要注 ...
- java泛型中的super和extend
List<? extend Fruit> list=new ArrayList<>(); 解释为:集合中元素是继承自Fruit,究竟是何种类型,编译器也无法判定. 如果要从集 ...
- 2016-06-13:NAT原理
参考资料 udp打洞( NAT traversal )的方法介绍 UDP打洞原理
- jQuery最佳编程实践
加载jQuery 1.坚持使用CDN来加载jQuery,这种别人服务器免费帮你托管文件的便宜干嘛不占呢.点击查看使用CDN的好处,点此查看一些主流的jQuery CDN地址. <script t ...