题目连接：
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 mi 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 ma - mb, 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, ..., mn (0 ≤ mi ≤ 2·104), where mi 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
2
1 6
Sample Output
2
Hint
题意
有n个机器，每个机器的任务时间是a[i]，你的任务是在最少的操作，使得所有Maxa[i]-Mina[i]最小。每次操作是可以使得某一个a[i]-1，某一个a[i]+1。
题解：
很显然，最后结果一定都在平均值位置，我们也可以比较轻松得到，最后究竟是多少个数为平均值，多少个数为平均值+1。所以我们直接贪心就好了。
小于平均值的就直接拉上去，大于平均值的就减下去就好了。代码
#includeusing namespace std;int a[100005];int main(){ int n;scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); sort(a+1,a+1+n); long long sum = 0; for(int i=1;i<=n;i++) sum+=a[i]; int p = sum / n; int n1 = n - sum%n; int ans = 0; for(int i=1;i<=n1;i++) { if(p<=a[i])break; ans+=p-a[i]; } for(int i=n1+1;i<=n;i++) { if(a[i]>p)break; ans+=p+1-a[i]; } printf("%d\n",ans);}