S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7262 Accepted Submission(s): 3074
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.
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
题意：m堆石子 玩家每次可以从某一堆中取出si个石子 不能取则输
题解：初步学习sg函数 sg[i]为 i的后继状 的sg值中 没有出现过的非负最小值。
sg异或值为0则后手胜
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 #include<bits/stdc++.h> 8 #include<map> 9 #include<set> 10 #include<cmath> 11 #include<queue> 12 #include<bitset> 13 #include<math.h> 14 #include<vector> 15 #include<string> 16 #include<stdio.h> 17 #include<cstring> 18 #include<iostream> 19 #include<algorithm> 20 #pragma comment(linker, "/STACK:102400000,102400000") 21 using namespace std; 22 #define A first 23 #define B second 24 const int mod=1000000007; 25 const int MOD1=1000000007; 26 const int MOD2=1000000009; 27 const double EPS=0.00000001; 28 typedef __int64 ll; 29 const ll MOD=1000000007; 30 const int INF=1000000010; 31 const ll MAX=1ll<<55; 32 const double eps=1e-8; 33 const double inf=~0u>>1; 34 const double pi=acos(-1.0); 35 typedef double db; 36 typedef unsigned int uint; 37 typedef unsigned long long ull; 38 int k; 39 int sg[10005]; 40 int a[105]; 41 int flag[105]; 42 int q,m,exm; 43 void init() 44 { 45 sg[0]=0; 46 for(int i=1;i<=10000;i++) 47 { 48 memset(flag,0,sizeof(flag)); 49 for(int j=1;j<=k;j++) 50 { 51 if(i-a[j]>=0) 52 { 53 flag[sg[i-a[j]]]=1; 54 } 55 } 56 for(int j=0;;j++) 57 { 58 if(flag[j]==0){ 59 sg[i]=j; 60 break; 61 } 62 } 63 } 64 } 65 int main() 66 { 67 while(scanf("%d",&k)!=EOF) 68 { 69 if(k==0) 70 break; 71 for(int i=1;i<=k;i++) 72 scanf("%d",&a[i]); 73 init(); 74 scanf("%d",&q); 75 for(int i=1;i<=q;i++) 76 { 77 scanf("%d",&m); 78 int ans=0; 79 for(int j=1;j<=m;j++) 80 { 81 scanf("%d",&exm); 82 ans^=sg[exm]; 83 } 84 if(ans==0) 85 printf("L"); 86 else 87 printf("W"); 88 } 89 printf("\n"); 90 } 91 return 0; 92 }
