D - Magic Powder - 1 codeforces 670d1

D - Magic Powder - 1

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.

Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value a_i — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use alln ingredients.

Apollinaria has b_i gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.

Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.

Input

The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.

The second line contains the sequence a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.

The third line contains the sequence b₁, b₂, ..., b_n (1 ≤ b_i ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.

Output

Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.

Sample Input

Input

3 1
2 1 4
11 3 16

Output

4

Input

4 3
4 3 5 6
11 12 14 20

Output

3

题意：有n中材料，每种材料有b克，他想做饼干，做1个饼干需要每种材料ai克，现在有k克魔法粉，这k克魔法粉可以变成任意一种材料，求最终最多做多少个饼干；

所有数据范围是10^3,这个的话我们可以直接暴力解决；

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<cctype>
#define max(a,b)(a>b?a:b)
#define min(a,b)(a<b?a:b)
#define INF 0x3f3f3f3f
typedef long long ll;

using namespace std;
#define N 1100

struct node
{
    int a,b;
}p[N];

int n,k;

int main()
{
    int i,ans;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d",&p[i].a);
        for(i=0;i<n;i++)
            scanf("%d",&p[i].b);

       ans=0;
       while(1)
       {
           for(i=0;i<n;i++)
           {
               if(p[i].b>=p[i].a)
                  p[i].b=p[i].b-p[i].a;

               else
               {
                   k=k-(p[i].a-p[i].b);
                   p[i].b=0;
               }

               if(i==n-1 && k>=0)
                ans++;
           }
           if(k<0)
               break;
       }
       printf("%d\n",ans);
    }
    return 0;
}