CodeForces - 808C-贪心-思维

本文解析了CodeForces上的一道题目,该题要求合理分配一大瓶酒至多个不同容量的酒杯中,需满足特定条件。文章分享了解题思路及代码实现,包括排序、贪心算法的应用。

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http://codeforces.com/problemset/problem/808/C
给定你n个酒杯的大小,
和你一大瓶酒 容量m,
要求1 每个杯至少放一半,奇数向上取整
2 容量大的杯子放的酒要比容量小的多
3 每个杯子里都要有,并且是整数性质。

是周赛的题,开始时交了十几遍,一直错在第8个样例。比赛完又自己开codeforce交了十几次。。终于发现了错误的原因竟然是两行的代码的顺序颠倒了,导致了一个失误。。。。
那就是在后来 把茶杯里剩下的给其他酒杯倒得时候,要先减去酒杯的容量,在更改cup的大小。。
总体是一个贪心,开始只是想的是每次 放一半,放完一半后,把剩下的一股脑的都放到最大的那个杯子里,没有考虑如果第一个又放满了怎么办。
后来看题解,发现了要从大到小一次再放。
(为什么让所有人都感到满意,这里是贪心的思想。)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn=3000;
struct Node
{   int u,v;
    int num;
};
int cmp1(Node a,Node b)
{  return a.u<b.u;//dadaopxiao
}
int cmp2(Node a,Node b)
{  return a.num<b.num;//dadaopxiao
}
int main()
{   int m;
   int w;
   Node v[maxn];
    scanf("%d%d",&m,&w);
      int sum=0;
      for(int i=0;i<m;i++)
      {  scanf("%d",&v[i].u);
         v[i].v=v[i].u/2+v[i].u%2;
         v[i].num=i;
          sum+=v[i].v;
      }
      if(sum>w){printf("-1\n");return 0;}
      sort(v,v+m,cmp1);
      w-=sum;
      int t=m-1;
      while(w>0)
      {   if(v[t].u-v[t].v<w)
            {   w-=(v[t].u-v[t].v);
                v[t].v=v[t].u;

            //w-=(v[t].u-v[t].v);
            t--;
                }
            else
                {v[t].v+=w;w=0;}
      }
        sort(v,v+m,cmp2);
      for(int x=0;x<=m-1;x++)
       cout<<v[x].v<<" ";
        //printf("%d\n",v[m-1].v);
    return 0;
}
### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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