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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)

## Chocolate Feast

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simple solution in c++

thanks broo

can you please explain how this works ?

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)