Stacking shelves (analog / thinking)

Ideas:

Solution 1:
because of the small data range, it can be simulated directly.

Method 2:
For each location, the number of required output T is necessary to add the number to the right, to the required output frequency and the left output (I prefix and calculation),
as a topic that can select any number of operating units moving, so take the T max is the answer

code (Method 2):

#include<bits/stdc++.h>
using namespace std;
const int maxm=105;
int a[maxm];
int sum[maxm];
signed main(){
    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        int tot=0;
        for(int i=1;i<=n;i++){
            cin>>a[i];
            tot+=a[i];
        }
        if(tot%n){
            cout<<-1<<endl;
            continue;
        }
        int av=tot/n;
        for(int i=1;i<=n;i++){
            sum[i]=sum[i-1]+av-a[i];
        }
        int ans=0;
        for(int i=1;i<=n;i++){
            int t=0;
            if(sum[i-1]>0)t+=sum[i-1];
            if(sum[n]-sum[i]>0)t+=sum[n]-sum[i];
            ans=max(ans,t);
        }
        cout<<ans<<endl;
    }
    return 0;
}
//http://oj.ecustacm.cn/problem.php?id=2003

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