Magic Powder - 1 CodeForces - 670D1 （优先队列）

最新推荐文章于 2020-05-16 00:45:23 发布

原创最新推荐文章于 2020-05-16 00:45:23 发布 · 418 阅读

0 ·

CC 4.0 BY-SA版权

STL 同时被 2 个专栏收录

16 篇文章

订阅专栏

优先队列

7 篇文章

订阅专栏

本文介绍了一个关于魔法粉和饼干制作的算法问题。主人公Apollinaria利用不同数量的食材和魔法粉来烘焙饼干，目标是计算出能制作的最大饼干数。通过优先考虑最紧缺的原料并使用魔法粉进行补充的方法，文章给出了一个有效的解决方案。

Magic Powder - 1

CodeForces - 670D1

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 all n 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 ningredients 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.

Example

Input

3 1
2 1 4
11 3 16

Output

Input

4 3
4 3 5 6
11 12 14 20

Output

Note

In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.

In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
#define INF 0x3f3f3f
int n,k;
struct cookie{
    int need;
    int has;
    bool operator < (const cookie &a)const{
        return (a.has / a.need) < (has / need);
    }
}cook[10010];
priority_queue<cookie>q;
int main(){
    int i,j;
    scanf("%d%d",&n,&k);
    for(i = 1; i <= n; i++){
        scanf("%d",&cook[i].need);
    }
    for(i = 1; i <= n; i++){
        scanf("%d",&cook[i].has);
        q.push(cook[i]);
    }
//    while(!q.empty()){
//        printf("%d ",q.top().has);
//        q.pop();
//    }
    while(k){
        cookie temp = q.top();
        q.pop();
        temp.has++;
        k--;
        q.push(temp);
    }
    int mincook = q.top().has/q.top().need;
    printf("%d\n",mincook);
    return 0;
}