For an integer and , unlimited denominations admit optimal greedy algorithm. In a minimum-count solution, no smaller denomination can occur times: carrying those stamps to the next denomination reduces their count by . Thus all counts below the maximum are base- digits, whose total value is at most . The maximum-denomination count is forced to be the quotient on division by , and repeating the argument forces every greedy remainder step.

Articles by others on the same topic (0)

There are currently no matching articles.