Codeforces 670D1. Magic Powder - 1 暴力

D1. Magic Powder - 1

time limit per test:

1 second

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

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 ai — 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 bi 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, ..., an (1 ≤ ai ≤ 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, ..., bn (1 ≤ bi ≤ 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.

Examples

input

3 1
2 1 4
11 3 16

output

4

input

4 3
4 3 5 6
11 12 14 20

output

3

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.

题目链接：http://codeforces.com/problemset/problem/670/D1

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题意：做一个饼干需要有n种材料，每种材料需要ai克。现在每种材料有bi克。还有k克神奇材料，可以代替那n种材料。1克神奇材料替换1克普通材料。

思路：按num[i]=bi/ai升序。暴力求出num[i]时需要得神奇材料。

 

代码：

for循环直接暴力：

#include<bits/stdc++.h>
using namespace std;
struct ingredient
{
    int a,b,num,sign;
} gg[];
int add[];
int cmp(ingredient x,ingredient y)
{
    if(x.num!=y.num) return x.num<y.num;
    else return x.sign<y.sign;
}
int main()
{
    int i,j,n,k;
    scanf("%d%d",&n,&k);
    for(i=; i<=n; i++)
        scanf("%d",&gg[i].a);
    for(i=; i<=n; i++)
    {
        scanf("%d",&gg[i].b);
        gg[i].num=gg[i].b/gg[i].a;
        gg[i].sign=gg[i].b%gg[i].a;
    }
    sort(gg+,gg+n+,cmp);
    gg[n+].num=gg[n].num+;
    gg[n+].sign=;
    int sum=,flag=;
    add[]=;
    for(i=; i<=n; i++)
    {
        sum+=gg[i].a;
        flag+=gg[i].sign;
        if(gg[i].num<gg[i+].num)
        {
            add[i]=add[i-]+sum*(gg[i+].num-gg[i].num)-flag;
            if(add[i]>=k)
            {
                k-=add[i-];
                cout<<gg[i].num+(k+flag)/sum<<endl;
                break;
            }
            flag=;
        }
        else add[i]=add[i-];
    }
    if(i>n)
    {
        k-=add[n];
        cout<<gg[n+].num+k/sum;
    }
    return ;
}

每次增加num就进行sort排序：

#include<bits/stdc++.h>
using namespace std;
int a[],b;
struct ingredient
{
    int a,b,num;
} gg[];
int cmp(ingredient x,ingredient y)
{
    return x.num<y.num;
}
int main()
{
    int i,n,k;
    scanf("%d%d",&n,&k);
    for(i=; i<n; i++)
        scanf("%d",&gg[i].a);
    for(i=; i<n; i++)
    {
        scanf("%d",&gg[i].b);
        gg[i].num=gg[i].b/gg[i].a;
    }
    sort(gg,gg+n,cmp);
    while(k>)
    {
        int sign=(gg[].num+)*gg[].a;
        if((sign-gg[].b)<=k)
        {
            k-=(sign-gg[].b);
            gg[].b=sign;
            gg[].num++;
        }
        else
        {
            gg[].b+=k;
            k=;
        }
        sort(gg,gg+n,cmp);
    }
    cout<<gg[].num<<endl;
    return ;
}