CodeForces 670D2 Magic Powder - 2

E - Magic Powder - 2
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
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670D2

Description

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.

Sample Input

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

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题意：有n中材料，每种材料有b克，他想做饼干，做1个饼干需要每种材料ai克，现在有k克魔法粉，这k克魔法粉可以变成任意一种材料，求最终最多做多少个饼干；

另一个所有数据范围变成10^9时我们可以想到，他最多做不超过2*(10^9)个饼干，所以我们可以二分搜索答案，一直到找到符合题意de饼干个数为止；

注意中间过程会爆int的所以用long long

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<cctype>
#include <map>
#define max(a,b)(a>b?a:b)
#define min(a,b)(a<b?a:b)
#define INF 0x3f3f3f3f
typedef long long ll;

using namespace std;
#define N 110000int n;
ll k,Max;
ll a[N],b[N];

ll Search(ll l,ll r)
{
while(l<=r)
{
ll mid=(l+r)/2;
ll sum=0;
for(int i=1;i<=n;i++)
{
if(b[i]<a[i]*mid)
sum=sum+(a[i]*mid)-b[i];
if(sum>k)
break;
}
if(sum==k)
return mid;
else if(sum<k)
l=mid+1;
else
r=mid-1;
}
return l-1;
}

int main()
{
while(scanf("%d%I64d",&n,&k)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%I64d",&a[i]);
for(int i=1;i<=n;i++)
scanf("%I64d",&b[i]);
Max=2*1e9+10;
ll ans=Search(0,Max);
printf("%I64d\n",ans);
}
return 0;
}