#include <iostream>

using namespace std;

int main()
{
    int k,n,s,i,g,r;
    cin>>k;
    s=0;
    cin>>n;
    int *a=new int[n];
    for (i=0;i<n;i++){
        cin>>a[i];
        s=a[i]+s;
        };
    g=s-(n*k);
    cout<<g;
    return 0;
}

consider this situation: when the number of the remaining cars until the current moment is less than k.
for example
5 3
3 6 5
the result is 1 not 0

попробуйте
5 5
0 0 0 0 0