poj2975--Nim】的更多相关文章

Nim Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5545   Accepted: 2597 Description Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or mor…
求处于必胜状态有多少种走法. if( (g[i]^ans) <= g[i]) num++; //这步判断很巧妙 // // main.cpp // poj2975 // // Created by New_Life on 16/8/12. // Copyright © 2016年 chenhuan001. All rights reserved. // #include <iostream> #include <stdio.h> #include <string.h&g…
DescriptionNim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or more stones from any single pile. Play ends when all the stones have been removed, at which…
Nim Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7279   Accepted: 3455 题目链接:http://poj.org/problem?id=2975 Description: Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s…
Nim Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5866   Accepted: 2777 Description Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or mor…
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>…
Nim poj-2975 题目大意:给定n堆石子,问:多少堆石子满足操作之后先手必胜. 注释:$1\le n\le 10^3$. 想法: 我们设M=sg(x1)^sg(x2)^...^sg(xn).其中,xi是第i堆石子个数. 如果sg(xi)^M<sg(xi),显然这堆石子满足题意. 最后,附上丑陋的代码... ... #include <iostream> #include <cstdio> #include <cstring> #include <al…
You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove th…
http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概率是在正面,各个卡牌独立.求把所有卡牌来玩Nim游戏,先手必胜的概率. (⊙o⊙)-由于本人只会在word文档里写公式,所以本博客是图片格式的. Code #include <cstdio> #include <cstring> #include <algorithm> u…
A Simple Nim Problem Description   Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed)…