complete solution:
int change(stack<int> denominations, int total){
int coin = denominations.pop();
int reminder = total % coin;
if(reminder == total) return 0; // in case you dont have a perfect
change
int numberOfCoins = (total-reminder)/coin;
if(reminder == 0) return numberOfCoins;
else return numberOfCoins + change(denominations, reminder);
}
On Oct 6, 11:01 am, ligerdave <david.c...@gmail.com> wrote:
> use mod recursively.
>
> total money(or reminder) mod denomination(big > small)
>
> On Oct 5, 7:13 pm, pre lak <pre.la...@gmail.com> wrote:
>
>
>
> > Hi all,
>
> > Pls help me with the solution to the following problem related to the coin
> > changing problem.
>
> > suppose that the available coins a ein the denominations c^0, c^1 , c^2...
> > c^k for some intgers c>1 and k>=1. show tht the greedy algorithm always
> > yeilds an optimal solution
>
> > thanks in advance
> > Preethi
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