2 seconds
256 megabytes
standard input
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.
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.
Print the minimum number of seconds required to balance the load.
2
1 6
2
7
10 11 10 11 10 11 11
0
5
1 2 3 4 5
3
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，每次可以增加其中一个，并同时减少另外一个，问最少多少次操作。
分析：显然平均数。
处理一下不能整除的情况。
/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define mk make_pair inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = ;
int n, arr[N];
LL sum, average, ans; inline void Input()
{
scanf("%d", &n);
for(int i = ; i < n; i++) scanf("%d", &arr[i]);
} inline void Solve()
{
for(int i = ; i < n; i++) sum += arr[i];
average = sum / n;
if(sum == average * n)
{
for(int i = ; i < n; i++) ans += abs(average - arr[i]);
ans /= ;
}
else
{
sort(arr, arr + n);
int number = sum - average * n;
for(int i = n - number; i < n; i++)
ans += abs(arr[i] - (average + ));
for(int i = ; i < n - number; i++)
ans += abs(arr[i] - average);
ans /= ;
}
cout << ans << endl;
} int main()
{
freopen("a.in", "r", stdin);
Input();
Solve();
return ;
}