题目来源：
http://codeforces.com/problemset/problem/1011/B
题目描述：
Natasha is planning an expedition to Mars for nn people. One of the important tasks is to provide food for each participant.
The warehouse has 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 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 nn and mm (1≤n≤1001≤n≤100, 1≤m≤1001≤m≤100) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers a1,a2,…,ama1,a2,…,am (1≤ai≤1001≤ai≤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 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 100100 participants and only 11 food package. In this case, the expedition can't last even 11 day.
大致题意：
有m袋食物，每个食物有一个编号，编号相同的就是同种食物，然后同一个人只能吃一种食物，不同的人可以相同或者不同的食物，每人每天吃1个单位的食物，问所有人最多能吃几天。。。
解题思路：
一开始想了半天，首先就是把同种食物的数量存在一个数组a中，只要从1天开始尝试d，然后扫一遍a，定义k=0，当作最多可以提高的人数，当ai>=d,k+=ai/d,直到扫完后，如果k>=n，那么就d++,直到不行为止，那么d就是最大天数。。
代码：
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,nx,o;
int a[109],x[109];
int f(int k)
{
int sum=0;
for(int i=1;i<=nx;i++)sum+=x[i]/k;
if(sum<n)return k-1;
else return f(k+1);
}
int main()
{
cin>>n>>m;
for(int i=1;i<=m;i++)cin>>o,a[o]++;
for(int i=1;i<=100;i++)
if(a[i])nx++,x[nx]=a[i];
cout<<f(1)<<endl;
return 0;
}