[Greedy] sharing Solitaire

Original title Portal

Thinking

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Really speaking, this question and contest a bit like building blocks, I started even confused, but the code did not do much worse, broadly consistent with the general idea.
Innocent greedy, as long as a number before a number is not as mean, put the number in the original basis with a number of former difference between the average and let the answer increment, the final output can answer ~~ ~

Code

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#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<map>
using namespace std;

int n,i,x,a[101],ans;

int main()
{
    cin>>n;
    for(i=1;i<=n;i++)
        cin>>a[i];
    //a[0]=a[n];
    for(i=1;i<=n;i++)
        x+=a[i];
    x/=n;
    for(i=0;i<=n;i++)
        a[i]-=x;
    for(i=2;i<=n;i++)
    {
        if(a[i-1]!=0)
            a[i]+=a[i-1],ans++;
    }
    
    cout<<ans;
    
    
    return 0;
}