codeforces Roud499Div2 B和C

B题数据不大暴力枚举答案即可

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

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

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

and m (1≤n≤100, 1≤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,…,am

(1≤ai≤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.

Examples

Input

Copy

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

Output

Copy

Input

Copy

100 1
1

Output

Copy

Input

Copy

2 5
5 4 3 2 1

Output

Copy

Input

Copy

3 9
42 42 42 42 42 42 42 42 42

Output

Copy

Note

In the first example, Natasha can assign type 1

food to the first participant, the same type 1 to the second, type 5 to the third and type 2 to the fourth. In this case, the expedition can last for 2 days, since each participant can get two food packages of his food type (there will be used 4 packages of type 1, two packages of type 2 and two packages of type 5

In the second example, there are 100

participants and only 1 food package. In this case, the expedition can't last even 1 day.

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<string.h>
using namespace std;
int n,m,x;
int a[107];
int main(){scanf("%d %d",&n,&m);for(int i=0;i<m;i++){scanf("%d",&x);a[x]++;}for(int i=1;i<=107;i++){int t=0;for(int j=1;j<=100;j++)t+=a[j]/i;if(t<n){printf("%d",i-1);return 0;}}return 0;
}

C题

要求最少携带多少燃料，那就让携带的燃料刚好够用，刚好够用就行。也就是最后降落地球的时候，燃料刚好用完了，飞船重量是初始中重量。由这个来倒着递推回去，即可知道刚开始的重量。

每次降落前或者起飞前的重量为x x-（x/系数）=落地后或者起飞后的重量

化简为（落地后的重量*系数）/（系数-1）=降落前的重量。

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=1007;
int n,a[maxn],b[maxn],flag;
double t;
int main(){scanf("%d %lf",&n,&t);for(int i=0;i<n;i++)scanf("%d",&a[i]);for(int i=0;i<n;i++)scanf("%d",&b[i]);double x=(t*b[0])/(b[0]-1);for(int i=n-1;i>=0;i--){if(a[i]-1<=0||b[i]-1<=0){flag=1;break;}x=(x*a[i])/(a[i]-1);if(i!=0)x=(x*b[i])/(b[i]-1);}if(flag)printf("-1\n");else printf("%.10lf\n",x-t);return 0;
}

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