Codeforces Round #499 (Div. 2) B 题二分

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Natasha计划组织一个火星探险队，需要为每个队员分配每日食物包，每包食物有特定类型。每个队员每天必须吃相同类型的食物。问题在于如何最大化探险队的持续天数，使得每天队员的食物类型都能满足。输入包含参与者数量和食物包数量，以及每个食物包的类型。输出是探险队能维持的最大天数。当无法规划至少一天的食物供应时，输出0。解答策略是通过二分查找来确定最大天数。

Planning The Expedition

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 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 jj Natasha should select his food type bjbj and each day jj-th participant will eat one food package of type bj. 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≤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,…,am (1≤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.

input

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

output

input

100 1
1

output

input

2 5
5 4 3 2 1

output

input

3 9
42 42 42 42 42 42 42 42 42

output

答案一定在0-种树最多的背包中取得，所以二分答案；

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=2e5+7;
const int INF=0x3f3f3f3f;
int cnt[150];
vector<int> v;
int n,k;
int a;
bool cmp(int a,int b)
{
    return a>b;
}
bool check(int x)
{
    int cnt=0;
    if(x==0) return true;
    for(int i=0;i<v.size();i++)
        cnt+=v[i]/x;
    return cnt>=n;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=k;i++)
        {
            scanf("%d",&a);
            cnt[a]++;
        }
    for(int i=1;i<=100;i++)
        if(cnt[i]) v.push_back(cnt[i]);
    sort(v.begin(),v.end(),cmp);
    if(v.size()==1)
    {
        cout<<*v.begin()/n<<endl;
        return 0;
    }
    int l=0;
    int r=*v.begin();
    int ans=0;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(check(mid)) l=mid+1,ans=mid;
        else r=mid-1;
    }
    cout<<ans<<endl;
}