codeforces 670D2. Magic Powder - 2

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 34 3 5 6
11 12 14 20

Output

3

题意：

制作一块饼干需要n种原料，每一种ai克。现在有每种原料分别bi克。而且你现在还有k克的魔法粉末，每一克粉末都可以当作任意一种原料一克使用。球最多能制作多少个饼干

思路：

直接二分饼干数+贪心即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int mod = 1000000007;
typedef long long LL;
#pragma comment(linker, "/STACK:102400000,102400000")
//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取
//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中cin
struct node
{
LL a, b;
}x[100005];
int main()
{
LL n, k;
int i, j;
while (cin >> n >> k)
{
for (i = 1; i <= n; ++i)
{
scanf("%lld", &x[i].a);
}
LL m = 0;
for (i = 1; i <= n; ++i)
{
scanf("%lld", &x[i].b);
m = max((x[i].b + k) / x[i].a, m);
}
LL L = 0, R = m;
LL mmax = 0;
//cout << L << R << endl;
while (L <= R)
{
LL mid = (L + R) / 2;
LL p = k;
LL ans = 0;

for (i = 1; i <= n; ++i)
{
if (p < 0)
{
ans = 1;
break;
}
if (x[i].b / x[i].a >= mid)
continue;
else
{
p -= mid * x[i].a - x[i].b;
}
if (p < 0)
{
ans = 1;
break;
}
}
if (ans)
{
R = mid - 1;
}
else
{
mmax = max(mmax, mid);
L = mid + 1;
}
//cout << mid <<" "<<mmax << endl;
}
printf("%lld\n", mmax);

}
}
/*
3 2
2 1 411 3 16

*/