[原创]codeforces 723C. Polycarp at the Radio [模拟]【杂类】

2016-10-04 18:34:39 Tabris_ 阅读数：313

博客爬取于2020-06-14 22:43:06
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52734843

题目链接:http://codeforces.com/contest/723/problem/C
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C. Polycarp at the Radio
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

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, …, an (1 ≤ ai ≤ 109), where ai 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 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.

Examples
input
4 2
1 2 3 2
output
2 1
1 2 1 2

input
7 3
1 3 2 2 2 2 1
output
2 1
1 3 3 2 2 2 1

input
4 4
1000000000 100 7 1000000000
output
1 4
1 2 3 4

Note
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.

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题目大意 ：
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。

解题思路：
就是模拟这个过程就行了

首先明确的是要想1-m之间的数中出现次数最少的尽量大那就是n/m个 （说白了就是匀乎匀乎。。。）

先记录一下1~m中每个数出现的次数 最后跟n/m比一下
然后就知道了 需要把改动多少个数了
我这里用另一个数组存储下改动的数

改动的 时候首先先把大于m的数给改成1~m个数
如果最后还有数需要改动就 在遍历一下数组 先统计下每个数的出现次数 只有出现次数==n/m之后才能改动这个数 这样才能保证1~m中每个数最少出现n/m个 然后注意下细节就能AC了、、

附本题代码
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# include<bits/stdc++.h>
using namespace std;

# define lalal puts("*****");
# define abs(a)  (a)>0?(a):-(a)
# define INF 0x1f1f1f1f
# define pb push_back

typedef long long int LL;
/***********************************/

int a[2222];
int b[2222];
int vis[2222];

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=m;i++) b[i]=0;

        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]<=m)   b[a[i]]++;
        }

        int sum=0,len=0,len1=0;
        for(int i=1;i<=m;i++)
            if(b[i]<n/m)
              for(int j=b[i];j<n/m;j++)  vis[len++]=i;

        for(int i=0;i<n&&len!=len1;i++)
            if(a[i]>m) a[i]=vis[len1++];

        for(int i=1;i<=m;i++) b[i]=0;

        for(int i=0;i<n&&len!=len1;i++)
        {
            if(b[a[i]]<n/m) b[a[i]]++;
            else      a[i]=vis[len1++];
        }

        printf("%d %d\n",n/m,len);
        for(int i=0;i<n;i++)
        {
            if(i) printf(" ");
            printf("%d",a[i]);
        }
        puts("");
    }
    return 0;
}