CodeForces - 609C Load Balancing

最新推荐文章于 2025-04-19 08:00:00 发布

原创最新推荐文章于 2025-04-19 08:00:00 发布 · 278 阅读

0 ·

CC 4.0 BY-SA版权

online judge CodeForces 同时被 2 个专栏收录

82 篇文章

订阅专栏

其他暴力

16 篇文章

订阅专栏

本文介绍了一个简单的服务器负载均衡问题及其实现方法。通过重新分配任务，使得各服务器间的负载尽可能均匀，最小化最忙服务器与最闲服务器之间的差异。

Load Balancing

题目描述

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.

输入

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.

输出

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

样例

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

题意

很简单的一个水题，因为是变成最平均，那么差值最大为一

AC代码

#include <bits/stdc++.h>

using namespace std;

#define LL long long
const int MAXN = 1e5+10;

int arr[MAXN];

int main() {
    int n;
    cin >> n;
    LL sum = 0;
    for(int i = 0; i < n; i++) {
        cin >> arr[i];
        sum += arr[i];
    }
    int ans = (int)sum/n;
    int maxx, minx;
    maxx = minx = ans;
    if(sum%n) maxx += 1;
    int z = 0, zz = 0;
    for(int i = 0; i < n; i++) {
        if(maxx < arr[i]) z += (arr[i]-maxx);
        else if(minx > arr[i]) zz += (minx-arr[i]);
    }
    cout << max(z, zz) << endl;
return 0;
}

/*


*/