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  1. Practice
  2. Algorithms
  3. Implementation
  4. Chocolate Feast
  5. Discussions

Chocolate Feast

  • hirenmangukiya86 + 2 comments

    simple solution in c++

    int main(){
    int t;
    cin >> t;
    for(int a0 = 0; a0 < t; a0++){
        int n;
        int c;
        int m;
        int t=0,w=0,l=0;
        cin >> n >> c >> m;
        t=n/c;
        w=n/c;
        while(w>=m)
        {
            t+=w/m;
            w=(w/m) + (w%m);
        }
        cout<<t<<endl;
    }
    return 0;
}
    • danyfamily22 + 0 comments

      thanks broo

    • agamyaji3 + 1 comment

      can you please explain how this works ?

      • vlomako + 0 comments

        So, how to solve the problem in two operators?

        Bobby uses money on the first iteration only. So, after 'n //= c' operator there are only two numbers in use: n - number of bars/wrappers Bobby has at the beginning, m - price of the promotional bar in wrappers.

        We can get rid of iterations when understanding that the chockolate bar "costs" not m wrappers, but m-1, because Bobby has one additional wrapper for every m of them. So the promotional bars number can be calculated without 'for/while' operator, simply dividing wrapper number by m-1.

        But Bobby can't spend all his wrappers. At least one wrapper remains. Thus the final formula is

        result = n [bars count bought for money] + (n-1)[wrappers count except one wrapper] // (m-1)