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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 100000000011000000000

output

2000000000

input

10 11000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 10000000001 1 1 1 1 1 1 1 1 1

output

0

input

3 12 1 411 3 16

output

4

input

4 34 3 5 611 12 14 20

output

3

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;}