codeforces670D2 Magic Powder - 2 （二分)

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题目链接 http://codeforces.com/problemset/problem/670/D2

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：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。

思路: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。

直接上代码：

#include <bits/stdc++.h>  using namespace std;  __int64 a[111111],b[111111];  const int maxn= 2 * 1e9 + 2;         //最多可以做这么多个蛋糕  __int64 n,k;  __int64 search(__int64 l,__int64 r)  {        __int64 mid,sum;        while(l <= r)      {            mid = ( l + r) / 2 ,sum = 0;            for (int i = 0 ; i < n ; i++ )       //遍历每种材料是否满足          {                sum += a[i] * mid - b[i] > 0 ? a[i] * mid - b[i] : 0 ;                if( sum > k)                  break;           }            if(sum == k)              return mid;            else if(sum < k)              l = mid + 1;            else               r = mid - 1;        }        return r;    }   int main()   {            while (~scanf("%I64d %I64d",&n,&k))      {          for (int i = 0 ; i < n ; i++)          {              scanf("%I64d",&a[i]);          }          for (int i = 0 ; i < n ; i++ )          {              scanf("%I64d",&b[i]);          }          printf("%I64d\n",search(1,maxn));      }      return 0;  }

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