Codeforces Round #350(Div.2)(D) 우선 순위 큐

D1. Magic Powder - 1
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.
Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.
Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.
Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.
Input
The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), 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 ≤ 1000), 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

3 1
2 1 4
11 3 16

output

4

input

4 3
4 3 5 6
11 12 14 20

output

3

Note
In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.
In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.
제목: 과자를 만들려면 과자마다 m가지의 다른 원료가 필요하고, 원료마다 a[i]가 필요합니다. 당신은 현재 원료마다 b[i]가 있습니다. 당신은 현재 K개의 마법가루가 있습니다. 마법가루는 어느 단위의 원료로 바뀔 수 있습니다. 최대 몇 개의 과자를 만들 수 있습니까?
문제풀이: 분명히 매번 b[i]/a[i]의 가장 작은 그것에 마법가루를 넣고 우선 대기열 유지보수를 사용한 다음에 가장 작은 b[i]/a[i] 출력을 찾아낸다.

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<set>  
#include<queue>  
#include<string>  
#include<bitset>  
#include<utility>  
#include<functional>  
#include<iomanip>  
#include<sstream>  
#include<ctime>  
using namespace std;

#define inf int(0x3f3f3f3f)  
#define mod int(1e9+7)  
typedef long long LL;
enum{ N = int(2e5 + 10) };
int a[N], b[N], c[N];
struct point
{
	int x, y;
	int num;
	bool operator<(const point& bb)const
	{
		return num > bb.num;
	}
	point(int _x, int _y, int _num) :x(_x), y(_y), num(_num){}
	point(){}
};

priority_queue<point>q;
int main()
{
#ifdef CDZSC  
	freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);
	int _time_jc = clock();
#endif  
	int n, k;
	while (~scanf("%d%d", &n, &k))
	{
		while (!q.empty())q.pop();
		for (int i = 0; i < n; i++)
		{
			scanf("%d", a + i);
		}
		for (int i = 0; i < n; i++)
		{
			scanf("%d", b + i);
		}
		for (int i = 0; i<n; i++)
		{
			q.push(point(a[i], b[i], b[i] / a[i]));
		}
		point top;
		while (k>0)
		{
			top = q.top(); q.pop();
			top.y += 1;
			k--;
			top.num = top.y / top.x;
			q.push(top);
		}
		int ans = inf;
		while (!q.empty())
		{
			ans = min(ans, q.top().num);
			q.pop();
		}
		printf("%d
", ans);
	}
	return 0;
}