map迭代器有点好用

Natasha is planning an expedition to Mars for $n$

people. One of the important tasks is to provide food for each participant.

The warehouse has $m$

$\mathrm{daily\; food\; packages.\; Each\; package\; has\; some\; food\; type\; ai}$

.

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 $j$

$\mathrm{Natasha\; should\; select\; his\; food\; type\; bj\; and\; each\; day\; j-th\; participant\; will\; eat\; one\; food\; package\; of\; type\; bj.\; The\; values\; bj}$

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 $n$

$\mathrm{and\; m\; (1\le n\le 100,\; 1\le m\le 100}$

) — the number of the expedition participants and the number of the daily food packages available.

The second line contains sequence of integers ${\mathrm{a1,a2,\dots ,am}}^{}$

$\mathrm{(1\le ai\le 100),\; where\; ai\; is\; the\; type\; of\; i}$

-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.

#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    map<int,int>mp;
    map<int,int>::iterator it;
    int n,m,b;
    int a[150],i;
    scanf("%d%d",&n,&m);
    for(i=0; i<=m-1; i++)
    {
        scanf("%d",&b);
        mp[b]++;
    }
    int t=0;
    for(it=mp.begin(); it!=mp.end(); it++)
    {
        a[t++]=it->second;

    }
    sort(a,a+t,cmp);
    int maxx=1;
    int flag;
    for(i=a[0]; i>=1; i--)
    {
        flag=0;
        int ge=0;
        for(int j=0; j<=t-1; j++)
            ge+=a[j]/i;
        if(ge>=n)
        {
            maxx=i;
            flag=1;
            break;
        }
    }
    if(flag==0)
        printf("0
");
    else
        printf("%d
",maxx);




}