CodeForces 670D2 Magic Powder 二分

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

题目大意：就是制作一个蛋糕需要n种材料，然后你有k克魔法粉，每克魔法粉可以代替任意一克的材料,ai代表制作一个蛋糕需要第i种材料多少克，bi代表你拥有第i个材料多少克，问做多可以做多少个蛋糕。

题解:二分查找可以制作多少个蛋糕，假如可以制作，那么每一种材料都必须充足。

#include<stdio.h>
const int maxn= 2 * 1e9 + 2;
__int64 n, t, a[100100], b[100100];
__int64 f(__int64 z,__int64 y)
{
    __int64 mid, sum;
    int i;
    while(z<=y)
    {
        mid=(z+y)/2;
        for(sum=0,i=1;i<=n;i++)
        {
            if(b[i]<a[i]*mid)
                sum += (a[i]*mid-b[i]);
            if(sum>t)
                break;
        }
        if(sum==t)
            return mid;
        else if(sum<t)
        z=mid+1;
        else
        y=mid-1;
    }
    return z-1;
}
int main()
{
    int i, j;
    scanf("%d%d", &n, &t);
    for(i=1;i<=n;i++)
        scanf("%d", &a[i]);
    for(j=1;j<=n;j++)
        scanf("%d", &b[j]);
    printf("%I64d
", f(0, maxn));

    return 0;
}