codeforces 603 C. Load Balancing 贪心

清疚 2022-08-18 15:22 25阅读 0赞

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 *m**i* 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 *m**a* - *m**b*, 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 *m*1, *m*2, ..., *m**n* (0 ≤ *m**i* ≤ 2·104), where *m**i* 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

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:

1.  move a task from server \#4 to server \#1 (the sequence *m* becomes: 2 2 3 3 5);
2.  then move task from server \#5 to server \#1 (the sequence *m* becomes: 3 2 3 3 4);
3.  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: (类比一下) 有n堆石子, 每堆有ai个石子, 每次可以从任意一堆转移一些石子到另外任意一堆, 要使最后石子最多的那堆与石子最小的那堆差的绝对值最小. 问最小转移的石子数.

分析: 如果石子总数能够整除n, 那么最小绝对值肯定是0, 反之, 肯定是1 ,因为如果是大于1的话, 那么再移动肯定能使绝对值变小.而且每一堆的石子数不是总数sum/n,就是sum/n+1,  那么我们就可以确定, 总和sum%n的余数肯定就是石子数为sum/n+1 的堆数. 那么把原先的数列排一下序, b\[i\]表示确定了应该的序列, 对应起来相减就是贪心的最优结果.

#include<bitset> //31ms
    #include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;
        while((c=getchar())<'0' || c>'9');
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res;
    }
    const int N=100100;
    
    int a[N],b[N];
    ll ans;
    int main()
    {
        int n=in(),sum=0;
    
        for(int i=0;i<n;i++)
        {
            a[i]=in();
            sum += a[i];
        }
        int ave=sum/n,res=sum%n;
        //列举出应该的序列,大的放在后面,已经排好序
        for(int i=n-1;i>=0;i--)
        {
            if(res) b[i]=ave+1,res--;
            else b[i]=ave;
        }
        sort(a,a+n); //对a数组进行排序
        for(int i=0;i<n;i++)
        {
            ans+=abs(a[i]-b[i]);  //对应的进行相减,保证总数最小
        }
        cout<<(ans>>1);
        return 0;
    }