Magic Powder CodeForces - 670 D2 二分法_codeforces magic二分题解-优快云博客

探讨在有限的魔法粉和各种原料下，如何通过算法计算出最多能制作多少个饼干。采用二分查找法优化解决方案，适用于大规模数据处理。

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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

题目链接

http://codeforces.com/problemset/problem/670/D2

题目大意

做饼干需要N种不同的材料，已有材料若干，和k份万能材料，问最多能做出多少个饼干。

数据范围

(1 ≤ n ≤ 100 000, 1 ≤ k ≤ 1e9)
(1 ≤ ai ≤ 1e9)
(1 ≤ bi ≤ 1e9)

解题思路

与前一题相比，数据更大。前一题数据小，可以用贪心得到答案。而这一题数据较大，用二分查找答案，复杂度更小，效率要高
（前一题链接）
http://codeforces.com/contest/670/problem/D1

解决代码

#include<cstdio>
typedef long long ll;
ll a[100005],b[100005];
int main()
{
ll n,k,l = 0,r = 2000000001,mid,sum,ans = -1;
scanf("%lld %lld",&n,&k);
for(int i = 1;i <= n;i++){
	scanf("%lld",&a[i]);
	}
	for(int i = 1;i <= n;i++){
		scanf("%lld",&b[i]);
		}
		while(l <= r){
			mid = (l + r)/2;
			sum = 0;
			for(int i = 1;i <= n;i++){
			if(b[i] < mid * a[i]) sum += mid * a[i] - b[i];
			if(sum > k) {
				break;
				}
			}
			if(sum > k)
			r = mid - 1;
			else if(sum == k){
				ans = mid;
				break;
			}
			else l = mid + 1;
		}	
			if(ans == - 1) ans = l - 1;
		
			printf("%lld\n",ans);
return 0;
}