codeforce之Magic Powder

最新推荐文章于 2017-01-17 11:04:16 发布

luckyu1

最新推荐文章于 2017-01-17 11:04:16 发布

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文章标签： codeforce

本文链接：https://blog.youkuaiyun.com/luckyu1/article/details/51582335

本文介绍了一个关于烘焙饼干的算法问题，通过使用二分查找法来确定最大可烘焙饼干数量。问题考虑了原料限制及魔法粉末的辅助作用，采用C++实现。

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题目:

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.

解答：

首先很很明显这是一个对结果进行二分的题目，

注意要用long long int，同时在判断目标答案是否可行的时候用减法不要用加法，否则可能会溢出

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#define ll long long int
using namespace std;

struct vege {
	ll id;
	ll per;
	ll total;
	ll alre;
	vege() {
	}
};

ll cost(vector<vege> &ingr, int tar,ll k) {
	ll res = 0;
	ll size = ingr.size();
	for (int i = 0; i < size; ++i) {
		if (k < 0)
			return false;
		k = k - (tar *ingr[i].per - ingr[i].total <= 0 ? 0 : tar *ingr[i].per - ingr[i].total);
	}
	if (k < 0)
		return false;
	return true;
}
static bool comp(vege A, vege B) {
	return A.alre < B.alre;
}
int main()
{
	ll n, k;
	cin >> n >> k;
	vector<vege> ingr(n);
	ll per;
	for (int i = 0; i < n; ++i) {
		cin >> per;
		ingr[i].id = i;
		ingr[i].per = per;
	}
	for (int i = 0; i < n; ++i) {
		cin >> per;
		ingr[i].total = per;
		ingr[i].alre = per / ingr[i].per;
	}
	//sort(ingr.begin(), ingr.end(), comp);
	int l = 0;
	int r = ~(1 << 31);
	int res = 0;
	while (l <= r) {
		int mid = l + (r - l) / 2;
		bool co = cost(ingr, mid, k);
		if (co) {
			l = mid + 1;
			res = mid;
		}
		else {
			r = mid - 1;
		}
	
	}
	cout << res << endl;
	//system("pause");
}