Magic Powder - 2

最新推荐文章于 2025-08-12 08:53:30 发布

爱上键盘的小哥哥

最新推荐文章于 2025-08-12 08:53:30 发布

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分类专栏：二分与三分

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本文介绍了一个烘焙问题，通过使用二分查找算法确定最大烘焙数量。输入包括原料需求、现有原料及魔法粉末数量，输出为最大烘焙数量。

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, ..., a_n (1 ≤ a_i ≤ 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, ..., b_n (1 ≤ b_i ≤ 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.

Example

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

题意：相似题

思路：如果能制作x块蛋糕，则一定能制作x-1，x-2块蛋糕

二分做蛋糕的数量

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1000000000
using namespace std;
typedef long long ll;
ll a[100005],b[100005];
ll n,k;
int judge(ll x){ //判断能不能做成x个蛋糕 
	ll ans=k;
	for(ll i=0;i<n;i++){ 
		if(x>b[i]/a[i]){ 
			if(ans<a[i]*x-b[i]) return 0;
			else ans-=a[i]*x-b[i];
		}
	}
	return 1;
}
int main(){
	scanf("%lld%lld",&n,&k);
	for(ll i=0;i<n;i++)
	scanf("%lld",&a[i]);
	for(ll i=0;i<n;i++)
	scanf("%lld",&b[i]);
	ll l=0,r=2*MAX+10;
	ll cnt=0;
	while(l<=r){ //二分做成的蛋糕数 
		ll mid=(l+r)/2;
		if(judge(mid)) l=mid+1,cnt=mid;
		else r=mid-1;
	}
	printf("%lld\n",cnt);
	return 0;
}