codeforces 350 div2 D Magic Powder - 2 二分

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

0

input

3 1
2 1 4
11 3 16

output

4

input

4 3
4 3 5 6
11 12 14 20

output

3
思路：直接二分查找最大能制造的饼干数，把上界设的2*10^9+1；最后你可能相差1，判断一下；
　　　（d-1直接暴力模拟就好了）

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
ll a[100010];
ll b[100010];
ll check(ll x,ll gg,ll y)
{
    ll cha=0;
    for(ll i=1;i<=x;i++)
    {
        if(a[i]*gg>b[i])
        cha+=a[i]*gg-b[i];
        if(cha>y)
        return 0;
    }
    return 1;
}
int main()
{
    ll x,y,z,i,t;
    scanf("%I64d%I64d",&x,&y);
    for(i=1;i<=x;i++)
    scanf("%I64d",&a[i]);
    for(i=1;i<=x;i++)
    scanf("%I64d",&b[i]);
    ll st=0;
    ll en=2000000000;
    while(st<en)
    {
        ll mid=(st+en)>>1;
        if(check(x,mid,y))
        st=mid+1;
        else
        en=mid;
    }
    if(check(x,st,y))
    printf("%I64d\n",st);
    else
    printf("%I64d\n",st-1);
    return 0;
}