Codeforces Round #499 (Div. 2)（1011）

Natasha is planning an expedition to Mars for $\mathrm{nn\; people.\; One\; of\; the\; important\; tasks\; is\; to\; provide\; food\; for\; each\; participant.}$

The warehouse has $\mathrm{mm\; daily\; food\; packages.\; Each\; package\; has\; some\; food\; type\; aiai\; .}$

Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.

Formally, for each participant $\mathrm{jj\; Natasha\; should\; select\; his\; food\; type\; bjbj\; and\; each\; day\; jj\; -th\; participant\; will\; eat\; one\; food\; package\; of\; type\; bjbj\; .\; The\; values\; bjbj\; for\; different\; participants\; may\; be\; different.}$

What is the maximum possible number of days the expedition can last, following the requirements above?

Input

The first line contains two integers $\mathrm{nn\; and\; mm\; (1\le n\le 1001\le n\le 100\; ,\; 1\le m\le 1001\le m\le 100\; ) —\; the\; number\; of\; the\; expedition\; participants\; and\; the\; number\; of\; the\; daily\; food\; packages\; available.}$

The second line contains sequence of integers ${\mathrm{a1,a2,\dots ,ama1,a2,\dots ,am\; (1\le ai\le 1001\le ai\le 100\; ),\; where\; aiai\; is\; the\; type\; of\; ii\; -th\; food\; package.}}^{}$

Output

Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.

Examples

Input

Copy

4 10
1 5 2 1 1 1 2 5 7 2

Output

Copy

2

Input

Copy

100 1
1

Output

Copy

0

Input

Copy

2 5
5 4 3 2 1

Output

Copy

1

Input

Copy

3 9
42 42 42 42 42 42 42 42 42

Output

Copy

3

Note

In the first example, Natasha can assign type $\mathrm{11\; food\; to\; the\; first\; participant,\; the\; same\; type\; 11\; to\; the\; second,\; type\; 55\; to\; the\; third\; and\; type\; 22\; to\; the\; fourth.\; In\; this\; case,\; the\; expedition\; can\; last\; for\; 22\; days,\; since\; each\; participant\; can\; get\; two\; food\; packages\; of\; his\; food\; type\; (there\; will\; be\; used\; 44\; packages\; of\; type\; 11\; ,\; two\; packages\; of\; type\; 22\; and\; two\; packages\; of\; type\; 55\; ).}$

In the second example, there are $\mathrm{100100\; participants\; and\; only\; 11\; food\; package.\; In\; this\; case,\; the\; expedition\; can\text{'}t\; last\; even\; 11\; day.}$

题意：有n个人去火星，有m斤食物，每个人只能吃一种食物，而且每天必须吃一斤食物，问n个人最多能活几天。

题解：由于数据量比较小，完全可以用暴力解决，一年一年往上加，不满足条件则找到最优解。

#include<stdio.h>
#include<string.h>
#include<stack>
#include<string.h>
#include<queue>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
const int MAXN=3e5+10;
map<int,int>::iterator it;
bool  cmp(int a,int b)
{
    return a>b;
}
int main()
{
    map<int,int >mp;
    int i,siz,n,rt,m,g=0,flag=0,ck=0;
    int ans[MAXN]= {0};
    int t[MAXN]= {0};
    int s[MAXN]= {0};
    scanf("%d%d",&n,&m);
    for(i=0; i<m; i++)
    {
        scanf("%d",&rt);
        mp[rt]++;
    }
    int r=n-mp.size();
    for(it=mp.begin(); it!=mp.end(); it++)
    {
        s[g++]=it->second;
    }
    sort(s,s+g,cmp);
    for(i=0; i<g; i++)
    {
        if(i==n)
            break;
        ans[ck++]=s[i];
    }
    for(; ck<n; ck++)
        ans[ck]=0;

    for(i=0; i<n; i++)
        t[i]=ans[i];
    for(i=1; i<=100; i++)
    {
        for(int j=0; j<n; j++)
        {
            if(ans[j]-i<0)
            {
                int k,cur=ans[j]-i;
                for(k=0; k<n; k++)
                {
                    int kk=ans[k]-i;
                    if(kk>=i)
                    {
                        ans[k]-=i;
                        ans[j]=0;
                        break;
                    }
                }
                if(k==n)
                {
                    flag=1;
                    break;
                }
            }

        }
        if(flag==1)
            break;
        for(int i=0; i<n; i++)
            ans[i]=t[i];

    }

    printf("%d
",i-1);


    return 0;
}

 当然这道题也可以二分查找

#include<stdio.h>
#include<string.h>
#include<stack>
#include<string.h>
#include<queue>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
const int MAXN=3e5+10;
int m,n,l,flag;
int str[MAXN];
map<int,int>::iterator it;
int  solve(int mid)
{
    int ans=0;
    for(int i=0;i<l;i++)
    {
        ans+=str[i]/mid;
    }
    if(ans>=m)
        flag=1;
    if(ans>=m)
        return 1;
    return 0;
}
int main()
{
    cin>>m>>n;
    int k;
    map<int,int>mp;
    for(int i=0;i<n;i++)
    {
        cin>>k;
        mp[k]++;
    }
    for(it=mp.begin();it!=mp.end();it++)
    {
        str[l++]=it->second;
    }
    int left=1,right=100,mid;
    while(left<=right)
    {
        mid=(left+right)/2;
        if(solve(mid))
        {
            left=mid+1;
        }
        else
        {
            right=mid-1;
        }
    }
    if(!flag)
        right=0;
    cout<<right<<endl;
    return 0;
}