CodeForces 670D1&&D2

最新推荐文章于 2019-02-21 21:26:57 发布

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本文介绍了一个关于烘焙饼干的问题，使用二分搜索算法来确定最大可烘焙的饼干数量。通过给定的食材数量和魔法粉末，算法评估了在不同约束条件下的解决方案。

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

D2的数据量是1e6其他的没变
这个我自己没想出来..
还好没硬怼看了题解不然明年我也做不出来

他直接二分了1到1e9的答案
对每个中间值进行检测从1到n是否可行…..
讲道理这么暴力的方法确实…..
666

#include<iostream>
#include<cmath>
#include<queue>
#include<stack>
#include<algorithm>
#include<cstdio>
using namespace std;
long long qw[100001],we[100001];
long long n,k;
int main()
{
    #define int long long 
    cin>>n>>k;
    for(int a=1;a<=n;a++)scanf("%I64d",&qw[a]);
    for(int a=1;a<=n;a++)scanf("%I64d",&we[a]);
    int tou=0,wei=2e9+7,jieguo;
    while(tou<=wei)
    {
        int zhong=(tou+wei)/2;
        int kk=k;
        int jiance=0,a;
        for(a=1;a<=n;a++)
        {
            if(zhong>we[a]/qw[a])
            {
                if(kk<qw[a]*zhong-we[a])break;
                else kk-=qw[a]*zhong-we[a];
            }
        }
        if(a>n)jiance=1;
        if(jiance)
        {
            jieguo=zhong;
            tou=zhong+1;
        }
        else 
        {
            wei=zhong-1;
        }
    }
    cout<<jieguo<<endl;
    return 0;
}