CodeForces 1011B Planning The Expedition (map迭代器)

最新推荐文章于 2018-08-15 23:02:05 发布

不想写代码了啊啊

最新推荐文章于 2018-08-15 23:02:05 发布

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本文链接：https://blog.youkuaiyun.com/shezjoe/article/details/81435405

本文介绍了一个关于为火星探险队员分配食物的问题，通过遍历不同天数来确定最多能活几天，涉及算法实现与数据结构应用。

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atasha 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

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

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.

神坑爹的的一道题，其实还是自己思维不行啊......菜得一逼啊（哭.jpg）

题意：n个人，m个口袋，每个口袋里边有一种类型的食物，一个人一天吃数量为1的食物，每个人从始至终只能吃一种类型的食物，问最多能活几天。

思路：遍历一边能存活的天数，看第i天能活下来j个人，当j<n的时候，说明活不过i天，最多只能存活i-1天。注意特判一下n>m的情况。

这道题学到了map迭代器的一点点用法，美滋滋。

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<map>
using namespace std;
typedef long long ll;
map<int,int> mp;
int main()
{
    int n,m,i,x,ans,sum;
    map<int,int>::iterator it; //开一个迭代器

    scanf("%d %d",&n,&m);
    for(i=1;i<=m;i++)
    {
        scanf("%d",&x);
        mp[x]++;
    }
    if(n>m)
    {
        printf("0\n");
    }
    else
    {
        for(i=1;;i++)
        {
            sum=0;
            for(it=mp.begin();it!=mp.end();it++) //遍历map
            {
                sum+=(it->second)/i; //it->first表示键，it->second表示值
            }
            if(sum<n)
                break;
        }
        printf("%d\n",i-1);
    }
    return 0;
}