Codeforces Round #375 (Div. 2) C Polycarp at the Radio-优快云博客

本文链接：https://blog.youkuaiyun.com/destiny546/article/details/52857441

本文介绍了一个关于音乐播放列表调整的问题，旨在通过改变播放列表使指定乐队的歌曲数量尽可能均匀分布，同时最小化所需更改次数。文章提供了一段C++代码实现，通过贪心算法策略解决了该问题。

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Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a₁, a₂, ..., a_n, where a_i 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 b_j 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, ..., b_m will be as large as possible.

Find this maximum possible value of the minimum among the b_j (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.

Input

The first line of the input contains two integers n and m (1 ≤ m ≤ n ≤ 2000).

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i is the performer of the i-th song.

Output

In the first line print two integers: the maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), where b_j 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.

水题，注意贪心顺序；

代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn=2005;
int vis[maxn],in[maxn],flag[maxn];
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&in[i]);
    }
    int all=n/m;
    for(int i=1;i<=n;i++)
    {
        int now=in[i];
        if(now<=m)
        {
            if(flag[now]<all)
            {
                flag[now]++;
                vis[i]=1;
            }
        }
    }
    int now=1,ans=0;
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            while(flag[now]>=all&&now<=m)
            {
                now++;
            }
            if(now>m)
                break;
            in[i]=now;
            flag[now]++;
            ans++;
        }
      // cout<<now<<endl;
    }
    cout<<all<<" "<<ans<<endl;
    for(int i=1;i<=n;i++)
    {
        printf("%d ",in[i]);
    }
    puts("");
    return 0;
}