S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3514 Accepted Submission(s): 1544
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
WWL
Source
Recommend
LL
人生中第一道SG函数题。。。。。
SG_DFS：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int k,s[111],t,m,a; int sg[11000]; int SG_dfs(int x) { if(sg[x]!=-1) { return sg[x]; } int i;bool vis[111]; memset(vis,false,sizeof(vis)); for(i=0;s { SG_dfs(x-s vis[sg[x-s } for(i=0;i<=10000;i++) { if(!vis } return sg[x]=i; } int main() { while(scanf("%d",&k)!=EOF&&k) { for(int i=0;i<k;i++) scanf("%d",s+i); sort(s,s+k); memset(sg,-1,sizeof(sg)); scanf("%d",&t); while(t--) { scanf("%d",&m); int XOR=0; while(m--) { scanf("%d",&a); XOR^=SG_dfs(a); } printf("%c",XOR?'W':'L'); } putchar(10); } return 0; } |
SG打表：
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int k,s[111],t,m,a; int sg[11000]; void getSG() { memset(sg,0,sizeof(sg)); bool flag[11000]; for(int i=1;i<=10000;i++) { memset(flag,false,sizeof(flag)); for(int j=0;j<k;j++) { if(s[j]>i) continue; flag[sg[i-s[j]]]=true; } for(int j=0;j<=10000;j++) { if(!flag[j]) { sg } } } } int main() { while(scanf("%d",&k)!=EOF&&k) { for(int i=0;i<k;i++) scanf("%d",s+i); getSG(); scanf("%d",&t); while(t--) { scanf("%d",&m); int XOR=0; while(m--) { scanf("%d",&a); XOR^=sg[a]; } printf("%c",XOR?'W':'L'); } putchar(10); } return 0; } |