http://poj.org/problem?id=2960 sg函数几乎是模板题. 调试代码的最大障碍仍然是手残在循环里打错变量名,是时候换个hydra产的机械臂了[超想要.jpg] #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<iostream> #include<map> using namespace std; ;…

题意: 有一块xi*Yi的矩形巧克力,Alice只允许垂直分割巧克力,Bob只允许水平分割巧克力.具体来说,对于Alice,一块巧克力X i * Y i,只能分解成a * Y i和b * Y i其中a + b = X i和a, b > 0.对于Bob,一块巧克力X i * Y i,只能分解成X i * a和X i * b其中a + b = Y i和a ,b > 0.(每次切割只能以整数单位来切,例如一个宽为3的巧克力,你垂直切只能切成一个1,2而不能切成两个1.5) 谁最后不能操作了,谁就输了…

S-Nim Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3694   Accepted: 1936 Description Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows: The starting position has a number of h…

预处理出SG函数,然后像普通nim一样做即可 #include<iostream> #include<cstdio> using namespace std; const int N=10005; int k,s[N],m,n,sg[N],v[N],ti,ans; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar(); } while(…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1056    Accepted Submission(s): 523 Problem Description Nim is a two-player mathematic game of strategy in which players take turn…

思路:直接打表找sg函数的值,找规律,没有什么技巧 还想了很久的,把数当二进制看,再类讨二进制中1的个数是必胜或者必败状态.... 打表: // #pragma comment(linker, "/STACK:102c000000,102c000000") #include <iostream> #include <cstdio> #include <cstring> #include <sstream> #include <str…

Code: #include<cstdio> #include<algorithm> #include<string> #include<cstring> using namespace std; #define maxn 10003 int step[maxn],SG[maxn],m,ans,l,a,k; bool vis[maxn]; int main(){ //freopen("input.in","r",std…

加强版的NIM游戏,多了一个操作,可以将一堆石子分成两堆非空的. 数据范围太大,打出sg表后找规律. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # inc…

Nim or not Nim? Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3032 Description Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps.…

题意: 有n个盒子,每个盒子可以放一定量的石头,盒子中可能已经有了部分石头.假设石头无限,每次可以往任意一个盒子中放石头,可以加的数量不得超过该盒中已有石头数量的平方k^2,即至少放1个,至多放k^2个. 思路: 跟常规nim的区别就是加了个限制“每次加的量不超平方”.盒子容量上限是100万,那么就不能直接计算SG了,会超时.sg打表后找规律.根据剩下多少个空位来决定sg值.都是0123456这样子递增的,碰到不能一次加满就变为0,然后继续递增,一直这样. 我的方案是,对于每个盒子大小,找到除了…