Educational Codeforces Round 3 C. Load Balancing

本文链接：https://blog.youkuaiyun.com/u014258433/article/details/50985394

本文探讨了如何通过重新分配任务来平衡多个服务器之间的负载，包括输入数据的解析、算法逻辑的实现以及输出结果的计算过程。通过示例说明了算法在不同场景下的应用，并强调了实现过程中需要注意的细节。

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time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 arem_i tasks assigned to thei-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 expressionm_a - m_b, wherea 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 thei-th server.

Output

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

Examples

Input

2
1 6

Output

Input

7
10 11 10 11 10 11 11

Output

Input

5
1 2 3 4 5

Output

Note

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.

改了好久才过的题，一定要注意细节。

#include <cstdio>
#include <iostream>
using namespace std;
int n,sum,ans,a[100001],ans2;
int main()
{
	scanf("%d",&n);
	for(int i=1;i <= n;i++)
	{
		scanf("%d",&a[i]);
		sum+=a[i];
	}
	int average=sum/n;
	for(int i=1;i <= n;i++)
	 if(a[i] > average) 
	 {
	 	ans+=a[i]-average;
	 	ans2++;
	 }
	if(sum % n != 0) printf("%d\n",ans-min(ans2,sum-average*n));
	else printf("%d\n",ans);
}