Internet Problem Solving Contest

IPSC 2013

Problem B – Boredom buster

Gillian is normally a very lively child. Most of the time she plays with her friends and tries to indulge in some mischief. But today is different, today Gillian woke up with the flu so she has to stay in bed – still and bored. To entertain her, her brother came up with the following game.

When Gillian has an integer x greater than 1, she can split it up into two positive integers y and z such that x = y + z. After performing this operation, her brother gives her y z hazelnuts. However, not all pairs of y,z are valid – there are some rules Gillian must comply with. These rules differ between the easy and hard subproblems; they are listed in the problem specification section.

Numbers that are obtained as a result of this operation can be also split up. At the beginning of the game, Gillian starts with a single integer n. She performs a series of operations described above until she is left with n copies of number 1. What is the maximum number of hazelnuts she can win if she chooses her moves wisely?

Problem specification – easy subproblem B1

For any x > 1 there is exactly one valid way of splitting:

Problem specification – hard subproblem B2

Gillian can pick any integer k satisfying 1 < k x and split up number x into y = x∕kand z = x −⌊x∕k.

Input specification

The first line of the input file contains an integer t 1000 specifying the number of test cases. Each test case is preceded by a blank line.

Each test case contains a single line with Gillian’s initial integer n > 1. You may assume that n 106 in the easy subproblem B1, and n 109 in the hard subproblem B2.

Output specification

For each test case, output a single line with the maximum number of hazelnuts.




This answer happens to be correct for both subproblems.