CodeForces - 670D2 Magic Powder - 2（二分）

最新推荐文章于 2017-11-15 21:29:36 发布

原创最新推荐文章于 2017-11-15 21:29:36 发布 · 541 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces #二分

2017暑假集训同时被 3 个专栏收录

16 篇文章

订阅专栏

题库_codeforces

14 篇文章

订阅专栏

搜索_二分搜索

4 篇文章

订阅专栏

本文介绍了一个关于原料配比与魔法粉末辅助烘焙的问题，通过二分查找算法确定最大可制作饼干数量的方法。该问题涉及到算法设计与实现，特别是二分搜索技巧的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

点击打开题目链接

D2. Magic Powder - 2

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The term of this problem is the same as the previous one, the only exception — increased restrictions.

Input

The first line contains two positive integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109) — the number of ingredients and the number of grams of the magic powder.

The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 109), 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 ≤ 109), 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

1 1000000000
1
1000000000

output

2000000000

input

10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1

output

input

3 1
2 1 4
11 3 16

output

input

4 3
4 3 5 6
11 12 14 20

output

题目大意：

有n种原料和k课魔法粉末，第二行为做一个饼干需要的每种原料数，第三行为每种原料总共有多少克。魔法粉末可以任意加到原料上，求最多可以做多少饼干。做的二分题还是少，当时没想到二分饼干数，一直超时样例。

思路：

二分饼干数求解。

附上AC代码：

#include<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;
const int maxn=100000+5;
ll n,k;
ll maxd;
ll sum;

struct nodes{
    ll need,have;
}node[maxn];

int main(){
    ios::sync_with_stdio(false);
    while(cin>>n>>k){
        sum=0;
        for(int i=1;i<=n;i++)
            cin>>node[i].need;
        for(int i=1;i<=n;i++){
            cin>>node[i].have;
            maxd=max((node[i].have+k)/node[i].need,maxd);
        }
        ll l=0,r=maxd;
        while(l<=r){
            ll mid=(l+r)/2;
            ll tmp=k;
            ll ans=0;
            for(int i=1;i<=n;i++){
                if(tmp<0){
                    ans=1;
                    break;
                }
                if(node[i].have/node[i].need>=mid)continue;
                else if(node[i].have/node[i].need<mid){
                    tmp-=mid*node[i].need-node[i].have;
                    if(tmp<0){
                        ans=1;
                        break;
                    }
                }
            }
            if(ans==1){//tmp<0,魔法粉末不够用
                r=mid-1;
            }
            else if(ans==0){
                sum=max(sum,mid);
                l=mid+1;
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}