Codeforces 609C Load Balancing 【水题】

C. Load Balancing

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 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 test(s)

input

2
1 6

output

2

input

7
10 11 10 11 10 11 11

output

0

input

5
1 2 3 4 5

output

3

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.

序列最平衡：min(max - min)。 

题意：给你n个元素的序列a[]，每次可以任选两个元素使得其中一个减1，另一个加1，付出代价为1。问你最小的代价使得序列最平衡。

思路：求出序列平衡时的上限Max和下限Min，显然小于下限的我们要提到Min，大于上限的我们要降到Max，而每次升和降是可以用代价1实现的。这样统计两边的代价，取最大值就好了。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
using namespace std;
int a[MAXN];
int main()
{
int n; Ri(n); int sum = 0;
for(int i = 0; i < n; i++)
Ri(a[i]), sum += a[i];
int Min, Max;
Min = Max = sum / n;
if(sum % n)
Max++;
int ans1 = 0, ans2 = 0;
for(int i = 0; i < n; i++)
{
if(a[i] < Min)
ans1 += Min - a[i];
else if(a[i] > Max)
ans2 += a[i] - Max;
}
Pi(max(ans1, ans2));
return 0;
}