HDOJ S-Nim 1536&POJ S-Nim 2960【求SG函数+Nim游戏】
2015-07-28 17:26
405 查看
S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5288 Accepted Submission(s): 2275
Problem 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 heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player's last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove
a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position.
This means, as expected, that a position with no legal moves is a losing position.
Input
Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0 < m ≤ 100) describing the
number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by a 0 on a line of its own.
Output
For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'. Print a newline after each test case.
Sample Input
2 2 5 3 2 5 12 3 2 4 7 4 2 3 7 12 5 1 2 3 4 5 3 2 5 12 3 2 4 7 4 2 3 7 12 0
Sample Output
LWW WWL
Source
Norgesmesterskapet 2004
Recommend
LL | We have carefully selected several similar problems for you: 1404 1517 1524 1729 1079
主要是求SG函数 然后利用Nim游戏求解方法求解 由此种游戏称Nim和
称一个游戏G 是一组游戏G1 G2 。。。Gn的和 G的每一步可抽象成:先选一个子游戏Gi 然后在Gi 中走一步。
如果gi是Gi的SG函数,则G的SG函数 g=g1^g2^g3^...^gn 对于G任意状态a=(a1,a2,,...,an),g(a)=g(a1)^g(a2)^...^g(an)。
此题将每次可取的数看成一个Gi 然后求解gi。最后进行Nim游戏。
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int sg[10010]; int s[110]; int k,m,l; int mex(int x) { if(sg[x]!=-1)return sg[x]; bool vis[110]; //一定要将vis数组定义到这里 memset(vis,false,sizeof(vis)); for(int i=0;i<k;i++){ int temp=x-s[i]; if(temp<0)break;//此处因为s[i]排好序的,所以后面temp的值肯定小于0,所以直接跳出就可以。 sg[temp]=mex(temp); vis[sg[temp]]=true; } for(int i=0;;i++){ if(!vis[i]){ sg[x]=i; break; } } return sg[x]; } int main() { while(scanf("%d",&k),k) { int res[110]; memset(res,0,sizeof(res)); memset(sg,-1,sizeof(sg)); sg[0]=0; for(int i=0;i<k;i++) scanf("%d",&s[i]); sort(s,s+k); scanf("%d",&m); int a; for(int i=0;i<m;i++){ scanf("%d",&l); for(int j=0;j<l;j++){ scanf("%d",&a); res[i]^=mex(a); } } for(int i=0;i<m;i++){ if(res[i]) printf("W"); else printf("L"); } printf("\n"); } return 0; }
相关文章推荐
- 简单研究Android View绘制二 LayoutParams
- The Independent JPEG Group's JPEG software Android源码中 JPEG的ReadMe文件
- iOS开发遇到的坑之三--使用asi框架在xcode下正常运行,但是打包时却不能进行网络访问
- C++中的顺序容器
- iOS中IMP指针的运用
- C++ 中dynamic_cast<>的使用方法
- IOS- 得到系统版本
- js判断用户的浏览设备是移动设备还是PC
- android学习笔记-第一天
- iOS多态
- MySQL入门级命令
- [转载] 使用 Twitter Storm 处理实时的大数据
- POJ 1456 Supermarket(贪心+并查集)
- centos xampp 隐藏phpmyadmin地址
- iOS开发使用nib进行界面设计并跳转
- 2d项目的内存优化
- 将 Shiro 作为应用的权限基础
- 大型网站架构之系列
- 计算机计算乘除法的原理
- VUA814-The Letter Carrier's Rounds(入门经典5-11,PE)