Codeforces723 C. Polycarp at the Radio (模拟)

来源:互联网 发布:生物统计 知乎 编辑:程序博客网 时间:2024/05/18 20:05

题目连接:http://codeforces.com/contest/723/problem/C


C. Polycarp at the Radio
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
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.

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 21 2 3 2
output
2 11 2 1 2 



input
7 31 3 2 2 2 2 1
output
2 11 3 3 2 2 2 1 



input
4 41000000000 100 7 1000000000
output
1 41 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.




题目大意:有n个数,要求将值为1到m中出现最小次数的数变成尽可能得多。


解题思路:题目意思有点感人……反正就是说1到m的数要至少出现n/m次,所以就把大于m的数压进一个队列里,然后把1到m中的出现次数大于n/m次的压进另一个队列中,然后就xjb搞吧……


/* ***********************************************┆  ┏┓   ┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃       ┃ ┆┆┃   ━   ┃ ┆┆┃ ┳┛ ┗┳ ┃ ┆┆┃       ┃ ┆┆┃   ┻   ┃ ┆┆┗━┓ 马 ┏━┛ ┆┆  ┃ 勒 ┃  ┆      ┆  ┃ 戈 ┗━━━┓ ┆┆  ┃ 壁     ┣┓┆┆  ┃ 的草泥马  ┏┛┆┆  ┗┓┓┏━┳┓┏┛ ┆┆   ┃┫┫ ┃┫┫ ┆┆   ┗┻┛ ┗┻┛ ┆************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <bitset>using namespace std;#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)#define pb push_back#define mp make_pairconst int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define LL long long#define ULL unsigned long long#define MS0(X) memset((X), 0, sizeof((X)))#define SelfType intSelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}#define Sd(X) int (X); scanf("%d", &X)#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())#define all(a) a.begin(), a.end()#define   mem(x,v)      memset(x,v,sizeof(x))typedef pair<int, int> pii;typedef pair<long long, long long> pll;typedef vector<int> vi;typedef vector<long long> vll;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")int a[2500];int t[2500];queue<int> q;queue<int> p;queue<int> res;int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);ios::sync_with_stdio(0);cin.tie(0);int n,m;n = read(), m = read();int cnt = n / m;MS0(t);for(int i=1;i<=n;i++){    scanf("%d",&a[i]);    if(a[i]>m)q.push(i);    else        {            t[a[i]]++;            if(t[a[i]]>cnt) p.push(i);        }}int sum = 0;    for(int i=1;i<=m;i++)    {        if(t[i]<cnt)        {            int tot = cnt -t[i];            while(tot--)res.push(i);        }    }while(!q.empty() && !res.empty())    {        int pos = q.front(); q.pop();        int val = res.front(); res.pop();        a[pos] = val;        sum++;    }while(!res.empty() && !p.empty())    {        int pos = p.front(); p.pop();        int val = res.front(); res.pop();        a[pos] = val;        sum++;    }    printf("%d %d\n",cnt,sum);    for(int i=1;i<=n;i++)    {        printf("%d",a[i]);        if(i==n)printf("\n");        else printf(" ");    }return 0;}


0 0