POJ 2975 Nim 尼姆博弈】的更多相关文章

题目大意:尼姆博弈,求先手必胜的情况数 题目思路:判断 ans=(a[1]^a[2]--^a[n]),求ans^a[i] < a[i]的个数. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h> #include<stdlib.h> #include<queue> #include<…
http://poj.org/problem?id=2975 题目始终是ac的最大阻碍. 问只取一堆有多少方案可以使当前局面为先手必败. 显然由尼姆博弈的性质可以知道需要取石子使所有堆石子数异或和为0,那么将某一堆a个石子变为a^异或和即可. a1^a2^a3^...^an=y; a1^a2^a3^...^an^y=0; #include<cstdio> #include<cstring> #include<algorithm> #include<cmath>…
LightOJ1253 :Misere Nim 时间限制:1000MS    内存限制:32768KByte   64位IO格式:%lld & %llu 描述 Alice and Bob are playing game of Misère Nim. Misère Nim is a game playing on k piles of stones, each pile containing one or more stones. The players alternate turns and…
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…
C. Industrial Nim time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output There are n stone quarries in Petrograd. Each quarry owns mi dumpers (1 ≤ i ≤ n). It is known that the first dumper of the…
题目大意:尼姆博弈,判断是否先手必胜. 题目思路: 尼姆博弈:有n堆各a[]个物品,两个人轮流从某一堆取任意多的物品,规定每次至少取一个,多者不限,最后取光者得胜. 获胜规则:ans=(a[1]^a[2] --^a[n]),若ans==0则后手必胜,否则先手必胜. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h>…
题目链接: https://cn.vjudge.net/problem/POJ-2234 题目描述: Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches fro…
这里是尼姆博弈的模板,前面的博弈问题的博客里也有,这里单列出来. 有N堆石子.A B两个人轮流拿,A先拿.每次只能从一堆中取若干个,可将一堆全取走,但不可不取,拿到最后1颗石子的人获胜.假设A B都非常聪明,拿石子的过程中不会出现失误.给出N及每堆石子的数量,问最后谁能赢得比赛. 例如:3堆石子,每堆1颗.A拿1颗,B拿1颗,此时还剩1堆,所以A可以拿到最后1颗石子. Input 第1行:一个数N,表示有N堆石子.(1 <= N <= 1000)第2 - N + 1行:N堆石子的数量.(1 &…
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…