S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2958 Accepted Submission(s):
1314
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
Recommend
LL
#include<stdio.h> #include<string.h> #include<stdlib.h> int f[101][101]; int arry[101]; int SG[10002]; int getsg(int num) { int i,tmp,hash[101]={0}; for(i=1;i<=arry[0];i++) { if(arry[i]>num) break; tmp=num-arry[i]; if(SG[tmp]==-1) SG[tmp]=getsg(tmp); hash[SG[tmp]]=1; } for(i=0;;i++) if(hash[i]==0) return i; } void px() { int i,j,k,tmp; for(i=1;i<arry[0];i++) { k=i; for(j=i+1;j<=arry[0];j++) if(arry[k]>arry[j]) k=j; tmp=arry[k]; arry[k]=arry[i]; arry[i]=tmp; } } int main() { int i,j,k,n,m,tmp,max; while(scanf("%d",&n)>0) { if(n==0)break; arry[0]=n; for(i=1;i<=n;i++) scanf("%d",&arry[i]); px(); scanf("%d",&m); for(i=1,max=0;i<=m;i++) { scanf("%d",&k); f[i][0]=k; for(j=1;j<=k;j++) { scanf("%d",&f[i][j]); if(f[i][j]>max) max=f[i][j]; } } memset(SG,-1,sizeof(SG)); for(i=1;i<=m;i++) { k=0; for(j=1;j<=f[i][0];j++) { tmp=f[i][j]; k=k^getsg(tmp); } if(k==0) printf("L"); else printf("W"); } printf("\n"); } return 0; }