Yeah! You are right and the description is easy. I sloved this problem with the same way. The following is my AC code :)

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int t,n,c,m;
    cin>>t;
    while(t--){
        cin>>n>>c>>m;
        int answer=0;
        // Computer answer(the total of chocolates Bob eats)
        // answer consists of two parts: directly buy with money and exchange with wrapper

        int answer1 = n / c; // the number of buying with money
        answer += answer1;

        int answer2 = 0; // the number of exchanging with wrapper

        int wrapper = answer1 / m; // the number of first exchange with wrapper
        int remain = answer1 % m; // the number of first remain wrapper
        answer2 += wrapper;

        int available = wrapper + remain;
        while (available / m != 0) {
            wrapper = available / m;
            remain = available % m;

            answer2 += wrapper;
            available = wrapper + remain;
        }

        answer += answer2;

        cout<<answer<<endl;
    }
    return 0;
}