标签：panel rem mem ora led class 参考 xor similar
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.
InputInput consists of a number of test cases.
For each test case: The rst 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.OutputFor 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
题解：可以当成多个NIM博弈，最终答案等于每个NIMA博弈结果的异或；(注意求SG函数时，不要每次都把vis数组清空，用一个t标记即可，每次改变标记，否则会超时)
参考代码：
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define clr(a,val) memset(a,val,sizeof a)
4 const int maxn=10010;
5 int num,f[maxn],ans;
6 int l,t,cas,SG[maxn],vis[maxn];
7 void GetSG(int x)
8 {
9 clr(SG,0);
10 int t=1;
11 for(int i=1;i<=x;++i)
12 {
13 for(int j=1;f[j]<=i&&j<=num;++j) vis[SG[i-f[j]]]=t;
14 for(int j=0;j<=x;j++) {if(vis[j]!=t) {SG[i]=j;break;}}
15 ++t;
16 }
17 }
18
19 int main()
20 {
21 while(~scanf("%d",&num) && num)
22 {
23 for(int i=1;i<=num;++i)scanf("%d",&f[i]);
24 sort(f+1,f+1+num);
25 GetSG(maxn-5);
26 scanf("%d",&cas);
27 while(cas--)
28 {
29 scanf("%d",&l);ans=0;
30 for(int i=1;i<=l;++i) scanf("%d",&t),ans^=SG[t];
31 if(!ans) printf("L");
32 else printf("W");
33 }
34 puts("");
35 }
36 return 0;
37 }
View Code
HDU1944 S-NIM(多个NIM博弈)
标签：panel rem mem ora led class 参考 xor similar
原文地址：https://www.cnblogs.com/songorz/p/10331512.html
