Catch That Cow

【文件属性】：
文件名称：Catch That Cow
文件大小：176KB
文件格式：RAR
更新时间：2013-11-04 19:38:03
上帝王牌 Catch That Cow 上帝王牌 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute * Teleporting: FJ can move from any point X to the point 2 × X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it? Input Line 1: Two space-separated integers: N and K Output Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow. Sample Input 5 17 Sample Output 4 Hint The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

【文件预览】：
Catch_That_Cow
----Catch_That_Cow.dsw(536B)
----Catch_That_Cow.opt(49KB)
----Catch_That_Cow.cpp(888B)
----Catch_That_Cow.dsp(5KB)
----StdAfx.cpp(301B)
----Catch_That_Cow.plg(2KB)
----StdAfx.h(769B)
----Debug()
--------StdAfx.obj(2KB)
--------Catch_That_Cow.ilk(180KB)
--------Catch_That_Cow.pch(199KB)
--------Catch_That_Cow.exe(184KB)
--------Catch_That_Cow.pdb(433KB)
--------vc60.idb(41KB)
--------Catch_That_Cow.obj(4KB)
--------vc60.pdb(52KB)
----Catch_That_Cow.ncb(49KB)
----ReadMe.txt(1KB)