HDU 1535 S-Nim（SG函数）
S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8729 Accepted Submission(s): 3660
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
题意
首先给出k，表示有几种每次取石子个数的集合，即给出123，每次可以取1,2,3个。然后给出询问次数m。每次询问给出n堆石子，然后询问当前状态是P还是N。
分析
SG函数。两种求法，都写了一遍。耗时间差不多
code
预处理
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 5 using namespace std; 6 7 const int N = 10000; 8 int sg[10100],f[1010]; 9 int k,n,m; 10 bool Hash[1010]; 11 12 void get_SG() { 13 memset(sg,0,sizeof(sg)); 14 for (int i=1; i<=N; ++i) { 15 memset(Hash,false,sizeof(Hash)); 16 for (int j=1; j<=k&&f[j]<=i; ++j) 17 Hash[sg[i-f[j]]] = true; 18 for (int j=0; j<=N; ++j) 19 if (!Hash[j]) {sg[i] = j;break;} 20 } 21 } 22 23 int main () { 24 while (~scanf("%d",&k) && k) { 25 for (int i=1; i<=k; ++i) 26 scanf("%d",&f[i]); 27 sort(f+1,f+k+1); 28 get_SG(); 29 scanf("%d",&m); 30 for (int i=1; i<=m; ++i) { 31 scanf("%d",&n); 32 int ans = 0; 33 for (int t,j=1; j<=n; ++j) { 34 scanf("%d",&t); 35 ans ^= sg[t]; 36 } 37 if (ans == 0) printf("L"); 38 else printf("W"); 39 } 40 puts(""); 41 } 42 return 0; 43 }
dfs记忆化搜索
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 6 int sg[10100],f[1010]; 7 int k,n,m; 8 9 int get_SG(int x) { 10 if (sg[x] != -1) return sg[x]; 11 bool Hash[110]; //不能再外面开 12 memset(Hash,false,sizeof(Hash)); 13 for (int i=1; i<=k; ++i) { 14 if (f[i] > x) break; 15 get_SG(x-f[i]); 16 Hash[sg[x-f[i]]] = true; 17 } 18 for (int i=0; ; ++i) 19 if (!Hash[i]) {sg[x] = i;break;} 20 return sg[x]; 21 } 22 int main () { 23 while (~scanf("%d",&k) && k) { 24 for (int i=1; i<=k; ++i) 25 scanf("%d",&f[i]); 26 sort(f+1,f+k+1); 27 memset(sg,-1,sizeof(sg)); 28 sg[0] = 0; 29 scanf("%d",&m); 30 for (int i=1; i<=m; ++i) { 31 scanf("%d",&n); 32 int ans = 0; 33 for (int t,j=1; j<=n; ++j) { 34 scanf("%d",&t); 35 ans ^= get_SG(t); 36 } 37 if (ans == 0) printf("L"); 38 else printf("W"); 39 } 40 puts(""); 41 } 42 return 0; 43 }
