Codeforces 670D. Magic Powder

Posted TOTOTOTOTZZZZZ

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Codeforces 670D. Magic Powder相关的知识,希望对你有一定的参考价值。

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.

这题还有大一点数据的版本,不过都是一样的

二分答案,每一次判断是O(n)的很好写

技术分享
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define SIZE 100005
#define MAXS 2000000005
LL n,k;
LL a[SIZE],b[SIZE];


bool ok(LL now){
  LL remain = k,oks = 1;
  for (int i=1;i<=n;i++){
    if (a[i] * now > b[i]){
      remain -= (a[i] * now - b[i]);
    }
    if (remain < 0){
      oks = 0; break;
    }
  }

  return oks;
}

LL search(LL l,LL r){
  // cout << l << " " << r << endl;
  if (l == r){
    return l;
  }
  if (l+1 == r){
    if (ok(r)){
      return r;
    }
    return l;
  }

  LL mid = l + (r - l) / 2;
  if (ok(mid)){
    return search(mid,r);
  }
  else
    return search(l,mid);
}

int main()
{
  // freopen("test.in","r",stdin);
  ios::sync_with_stdio(false);

  cin >> n >> k;

  for (int i=1;i<=n;i++){
    cin >> a[i];
  }
  for (int i=1;i<=n;i++){
    cin >> b[i];
  }

  LL ans = search(0,MAXS);
  cout << ans;
  return 0;
}
View Code

 

以上是关于Codeforces 670D. Magic Powder的主要内容,如果未能解决你的问题,请参考以下文章

CodeForces 670D Magic Powder

CodeForces 670A Holidays

CodeForces 670A Holidays

codeforces 628D. Magic Numbers 数位dp

[题解]CodeForces878 D - Magic Breeding

Educational Codeforces Round 8 D. Magic Numbers