hdu 1536 S-Nim(SG函数)
2017-10-14 12:14
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S-Nim
Problem DescriptionArthur 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.
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
思路:最基础的SG函数
注意:给出的S集合并不一定是单调递增的
SG函数学习博客:Strangedbly
代码:<
4000
/p>
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn=1e4+10; int sg[maxn],f[110],a[110][110]; bool vis[maxn]; int k; void Get_SG(int n) { memset(sg,0,sizeof(sg)); for(int i=1; i<=n; ++i) { memset(vis,false,sizeof(vis)); for(int j=1; j<=k&&f[j]<=i; ++j) vis[sg[i-f[j]]]=true; for(int j=0;; ++j) if(!vis[j]) { sg[i]=j; break; } } } int main() { while(scanf("%d",&k),k) { for(int i=1; i<=k; ++i) scanf("%d",&f[i]); sort(f+1,f+1+k); int m,n,maxx=0; scanf("%d",&m); for(int i=1; i<=m; ++i) { scanf("%d",&a[i][0]); for(int j=1; j<=a[i][0]; ++j) scanf("%d",&a[i][j]),maxx=max(maxx,a[i][j]); } Get_SG(maxx); for(int i=1; i<=m; ++i) { int ans=0; for(int j=1; j<=a[i][0]; ++j) ans^=sg[a[i][j]]; if(ans) printf("W"); else printf("L"); } puts(""); } return 0; }
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