题解:尼姆博弈,对于1至1000计算SG函数,每次取最小的前继值,SG值异或为0则为P-position. #include <cstdio> #include <cstring> using namespace std; int fbi[30]; int SG[1001]; int m,n,p; int main(){ fbi[0]=0,fbi[1]=1,fbi[2]=2; for(int i=3;fbi[i-1]<=1000;i++)fbi[i]=fbi[i-1]+fbi…

Fibonacci again and again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5088    Accepted Submission(s): 2126 Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的:F(1)=1;F(2)=2;…

Rabbit and Grass Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3864    Accepted Submission(s): 2919 Problem Description 大学时光是浪漫的,女生是浪漫的,圣诞更是浪漫的,但是Rabbit和Grass这两个大学女生在今年的圣诞节却表现得一点都不浪漫:不去逛商场,不去逛…

Climbing the Hill Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description Alice and Bob are playing a game called "Climbing the Hill". The game board consists of cells arranged vertically, as the…

http://acm.hdu.edu.cn/showproblem.php? pid=1849 简单的尼姆博弈: 代码例如以下: #include <iostream> #include <cstdio> using namespace std; int main() { int m,n,t; while(cin>>m,m) { int ans=0; for(int i=0; i<m; i++) { cin>>n; ans^=n; } if(ans==…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1907 Problem Description Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same c…

HDU.1850 Being a Good Boy in Spring Festival (博弈论 尼姆博弈) 题意分析 简单的nim 博弈 博弈论快速入门 代码总览 #include <bits/stdc++.h> #define nmax 105 using namespace std; int a[nmax]; int main() { int m; while(scanf("%d",&m) != EOF && m){ int ans = 0,…

题目链接: PKU:http://poj.org/problem? id=3480 HDU:http://acm.hdu.edu.cn/showproblem.php? pid=1907 Description Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to…

参考博客 先讲一下Georgia and Bob: 题意: 给你一排球的位置(全部在x轴上操作),你要把他们都移动到0位置,每次至少走一步且不能超过他前面(下标小)的那个球,谁不能操作谁就输了 题解: 当n为偶数的时候,假设当每个球都相互挨着没有间隙,那么两两一组,一组中前面那个走到哪,后面那个跟上就可以了,先手必输 如果球与球之间有间隙,那么俩俩球之间的距离可以当作尼姆博弈中取石子游戏中一堆石子的石子数,用尼姆博弈判断一下就可以了 可以说先手赢不赢光和两球之间的距离有关,如果俩俩球之间的的距离…

Fibonacci again and again HDU - 1848 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的: F(1)=1; F(2)=2; F(n)=F(n-1)+F(n-2)(n>=3); 所以,1,2,3,5,8,13……就是菲波那契数列. 在HDOJ上有不少相关的题目,比如1005 Fibonacci again就是曾经的浙江省赛题. 今天,又一个关于Fibonacci的题目出现了,它是一个小游戏,定义如下: 1.  这是一…