CodeForces 609C Load Balancing

最新推荐文章于 2018-04-25 21:37:56 发布

原创最新推荐文章于 2018-04-25 21:37:56 发布 · 410 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

本文介绍了一种通过重新分配任务来平衡服务器负载的算法，旨在最小化服务器间任务数量的差异，提供了一个具体的实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

C - Load Balancing

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 609C

Description

In the school computer room there are n servers which are responsible for processing several computing tasks. You know the number of scheduled tasks for each server: there are m_i tasks assigned to the i-th server.

In order to balance the load for each server, you want to reassign some tasks to make the difference between the most loaded server and the least loaded server as small as possible. In other words you want to minimize expression m_a - m_b, where a is the most loaded server and b is the least loaded one.

In one second you can reassign a single task. Thus in one second you can choose any pair of servers and move a single task from one server to another.

Write a program to find the minimum number of seconds needed to balance the load of servers.

Input

The first line contains positive number n (1 ≤ n ≤ 10⁵) — the number of the servers.

The second line contains the sequence of non-negative integers m₁, m₂, ..., m_n (0 ≤ m_i ≤ 2·10⁴), where m_i is the number of tasks assigned to the i-th server.

Output

Print the minimum number of seconds required to balance the load.

Sample Input

Input

2
1 6

Output

Input

7
10 11 10 11 10 11 11

Output

Input

5
1 2 3 4 5

Output

Hint

In the first example two seconds are needed. In each second, a single task from server #2 should be moved to server #1. After two seconds there should be 3 tasks on server #1 and 4 tasks on server #2.

In the second example the load is already balanced.

A possible sequence of task movements for the third example is:

move a task from server #4 to server #1 (the sequence m becomes: 2 2 3 3 5);
then move task from server #5 to server #1 (the sequence m becomes: 3 2 3 3 4);
then move task from server #5 to server #2 (the sequence m becomes: 3 3 3 3 3).

The above sequence is one of several possible ways to balance the load of servers in three seconds.

题意：n个数表示n个服务器每个服务器的任务数，然后你需要调配，确保每个服务器上的任务数相差不超过1。问你最少的步数。

思路：我们设这n个数的平均数为x，则这n个数的最终状态不外乎两种x或者x+1。

那么我们可以分别记录

1.a[i]<x的情况上a[i]需要增加的步数ans1。

2.a[i]>x+1的情况上需要减小的步数ans2。

然后取之中的较大者就可以了。

你问中间的值怎么办？当然是凉拌咯~你看看剩下的值是哪种值？

为什么取较大值？因为这两种情况都要满足！差的步数怎么办？当然是x来补咯~

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int MAXN=100000+5;
int a[MAXN];
int main()
{
    int n;
    scanf("%d",&n);
    int i;
    int sum=0;
    for(i=0;i<n;++i)
    {
     scanf("%d",&a[i]);
     sum+=a[i];
    }
  sort(a,a+n);
  int x=sum/n;
  int ans=0;
  int ans1=0;
  for(i=0;i<n;++i)
  {
      if(x>a[i])ans+=x-a[i];
      else if(a[i]>x+1)ans1+=a[i]-x-1;
  }
    ans=max(ans,ans1);
    printf("%d\n",ans);

    return 0;
}