UVa 11264 - Coin Collector

Summary
This problem can be solved in O(n) time using greedy approach. Given types of coins available, problem asks to find the maximum number of "different types" of coins, we can get in "one" withdraw. We have also been given the algorithm bank follows for withdrawing :

withdraw(X){ if( X == 0) return; Let Y be the highest valued coin that does not exceed X. Give the customer Y valued coin. withdraw(X-Y); }

Explanation
Now there are two cases why ? because as explained in withdraw(x) algorithm followed by bank, they will simply give us (i+1)th coin. In this case don't forget to update S to S+value of ith coin
 * First point to understand is we always get first and last(maximum) coin. how? ask any amount > maximum coin and you will get maximum valued coin. similarly ask for amount==minimum coin value and you will get minimum coin.
 * Now we have to decide for other coins (whether or not we can get them ?)
 * let's see how to decide about ith coin (1 'S'+(value of ith coin).

Use this approach to greedily decide about all i from 1 to n-1.

Gotchas

 * Types of coin in input are sorted in ascending order of values, so do not waste time in sorting.
 * minimum number of coins is 2, except the case when there is only one coin available.
 * We have to maximize and calculate the number of "different types" of coin in single withdraw. Amount to be withdrawn is irrelevant.

Implementations
Modify to handle when n = 1;
 * http://ideone.com/7AdROI

Optimizations
Optimizations here.

Input
8 6 1 2 4 8 16 32 6 1 3 6 8 15 20 7 1 5 9 74 111 121 159 10 1 2 3 4 5 6 7 8 9 10 5 1 2 4 8 15 8 1 5 9 17 25 33 42 100 16 1 2 4 17 58 69 125 254 478 1023 10000 145236 172589 172590 1000000 10000000 2 1 2

Output
6 4 5 2 4 5 14 2