CodeForces--609C --Load Balancing(水题)

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原文链接：http://www.cnblogs.com/playboy307/p/5273520.html

本文探讨了在给定服务器任务分配的情况下，如何通过每秒重新分配单个任务来平衡服务器负载，最小化最繁忙与最空闲服务器之间的任务数量差异。

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Load Balancing

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

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

Sample 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.

Source

Educational Codeforces Round 3

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num[1000010];
int main()
{
	int n;
	scanf("%d",&n); 
	int sum=0;
	int maxx,minn;
	for(int i=0;i<n;i++)
	{
		scanf("%d",&num[i]);
		sum+=num[i];
	}
	minn=maxx=sum/n;
	int ans1,ans2;
	ans1=ans2=0;
	if(sum%n) 
	maxx++;
	for(int i=0;i<n;i++)
	{
		if(num[i]<minn) ans1+=minn-num[i];
		else if(maxx<num[i]) ans2+=num[i]-maxx;
	}
	int s=max(ans1,ans2);
	printf("%d\n",s);
	return 0;
}

转载于:https://www.cnblogs.com/playboy307/p/5273520.html