Magic Powder - 1 CodeForces - 670D1 (优先队列)
2017-11-15 21:28
405 查看
Magic Powder - 1
CodeForces- 670D1
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 ningredients 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.
Example
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.
code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
#define INF 0x3f3f3f
int n,k;
struct cookie{
int need;
int has;
bool operator < (const cookie &a)const{
return (a.has / a.need) < (has / need);
}
}cook[10010];
priority_queue<cookie>q;
int main(){
int i,j;
scanf("%d%d",&n,&k);
for(i = 1; i <= n; i++){
scanf("%d",&cook[i].need);
}
for(i = 1; i <= n; i++){
scanf("%d",&cook[i].has);
q.push(cook[i]);
}
// while(!q.empty()){
// printf("%d ",q.top().has);
// q.pop();
// }
while(k){
cookie temp = q.top();
q.pop();
temp.has++;
k--;
q.push(temp);
}
int mincook = q.top().has/q.top().need;
printf("%d\n",mincook);
return 0;
}
相关文章推荐
- codeforces 670D1 Magic Powder - 1
- codeforces 670D1 - Magic Powder - 1
- D - Magic Powder - 1 codeforces 670d1
- CodeForces - 670D1 Magic Powder - 1 (模拟)
- CodeForces - 670D1 Magic Powder - 1 (模拟)
- CodeForces - 799B T-shirt buying —— 贪心 优先队列
- CodeForces 140C - New Year Snowmen(优先队列)
- CodeForces 670D2 Magic Powder - 2
- CodeForces - 864D Make a Permutation! 水题(用了优先队列)
- CodeForces - 670D2 Magic Powder - 2 (二分&模拟)
- CodeForces 3D Least Cost Bracket Sequence (贪心+优先队列)
- codeforces 754D Fedor and coupons【优先队列+贪心*好题】
- CodeForces 140C 贪心+优先队列
- 【Codeforces 747 C Servers 】+ 思维 或 优先队列
- Codeforces 767E 优先队列
- CodeForces 337B Preparing for the Contest(二分+贪心+优先队列)
- Magic Powder - 2 CodeForces - 670D2(二分)
- codeforces 913D Too Easy Problems (贪心+优先队列)
- Codeforces 721D Maxim and Array【贪心+优先队列+分类讨论】
- codeforces 670D2 - Magic Powder - 2