题目链接 A Sample Stone Game 题目大意:给定n,k,表示最初时有n个石头,两个人玩取石子游戏,第一个人第一次可以取1~n-1个石头,后面每个人最多可以拿走前面一个人拿走的个数的K倍,当有一个人可以一次性全部拿走时获胜.问两人都在不失误的情况下,先拿着有没有必胜局势.有的话求他第一次最少该取多少个. 思考过程: 首先讨论k=1的情况,我们可以把一个数n(石子的个数),写为二进制下的表示,那先者取走最后一个1,那后者必然不能取走比它高一位的1,那么先拿者一定会赢,当然如果n本来就…

题目: E - A simple stone game Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3922 Description After he has learned how to play Nim game, Mike begins to try another stone game which seems much eas…

先是题目,本来是第三次训练的题,在这特别提出来讲. 先是题目: E - A simple stone game Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit   Status   Practice   POJ 3922 Description After he has learned how to play Nim game, Mike begins to try anoth…

A simple stone game                                                                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                             …

A Simple Stone Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description After he has learned how to play Nim game, Bob begins to try another ston…

A Simple Stone Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1526    Accepted Submission(s): 346 Problem Description After he has learned how to play Nim game, Bob begins to try another…

A Simple Stone Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description After he has learned how to play Nim game, Bob begins to try another ston…

option=com_onlinejudge&Itemid=8&page=show_problem&problem=4342">题目链接:uva 1567 - A simple stone game 题目大意:给定K和N.表示一堆石子有N个.先手第一次能够取1~N-1个石子,取到最后一个石子的人胜利,单词每次操作时,取的石子数不能超过对手上一次取的石子数m的K倍. 问先手能否够必胜.能够输出最小的首次操作. 解题思路:这题想了一天,又是打表找规律.又是推公式的,楞是…

思路: 这就是K倍动态减法游戏,可以参考曹钦翔从“k倍动态减法游戏”出发探究一类组合游戏问题的论文. 首先k=1的时候,必败态是2^i,因为我们把数二进制分解后,拿掉最后一个1,那么会导致对方永远也取不完,我们可以拿到最后一个1. k=2的时候,必败态是斐波那契数列,因为任何一个整数n都可以写成两项斐波那契数的和,所以我们拿掉1,对方永远取不完高两位的数. k的时候我们必须构造数列,将n写成数列中一些项的和,使得这些被取到的项的相邻两个倍数差距>k 那么每次去掉最后一个1 还是符合上面的条件.设…

题目:http://poj.org/problem?id=3468   A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 85851   Accepted: 26685 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with…