Here is my c++ solution, you can watch the explanation here : https://youtu.be/retgbr5jSsg

int jumpingOnClouds(vector<int> c, int k) {
    int position = 0, energie = 100;
    do{
        position = (position + k) % c.size();
        energie -= 1 + c[position] * 2;
    }while(position != 0);
    return energie;
}

Here's a C solution: int jumpingOnClouds(int c_count, int* c, int k) { int e = 100, i=0, beg = c[0], end = -1; c[0] = 111;
while(end!=111){ if(c[i]==1){ e = e-2; } e--; if((i+k)>c_count-1) {
i = abs(i+k-c_count); } else{ i+=k; }
end = c[i]; } if(beg == 1){ e-=2; } return e; }

Here's a C# solution:

            int energy = 100;
    int cloudIndex = k % c.Count();
    if (c[cloudIndex] == 1)
    {
        energy -= 3;
    }
    else energy--;
    while (cloudIndex % c.Count() != 0)
    {
        cloudIndex += k;
        if (c[cloudIndex % c.Count()] == 1)
        {
            energy -= 3;
        }
        else energy--;
    }
    return energy;

Perl solution:

sub jumpingOnClouds {
    my ($c, $k) = @_;
    
    my $energy = 100;
    my $ind = 0;
    for (my $i = 0; $i < scalar(@$c); $i++) {
        $ind = ($k + $ind) % scalar(@$c);
        $energy--;
        $energy -= 2 if ($c->[$ind] == 1);
        last if ($ind == 0);
    }
    return $energy;
}

Java 8 Code

// Complete the jumpingOnClouds function below.

static int jumpingOnClouds(int[] c, int k) {
    int index =0;
    int e =100;
    for(int i =0;i<c.length;i++){
        index = (index+k)%c.length;
        if(c[index]==1){
            e = e-3;
        }else{
            e = e-1;
        }
        if(index==0){
            break;
        }
    }

  return e;
}