标签：after his set bee always ogr false bsp cout
S-Nim
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.
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
N推石头，每次只能从一堆中取S[i] 个石头
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int N = 110;
4 int s[N*N], SG[N*N], f[N];
5 int k, m, l, y, x;
6 int getSG(int n) {
7 int mex[101];
8 memset(mex, 0, sizeof(mex));
9 if(SG[n] != -1) return SG[n];
10 if(n - f[0] < 0) return SG[n] = 0;
11 for(int i = 0; i < k && f[i] <= n; i ++) {
12 mex[getSG(n - f[i])] = 1;
13 }
14 for(int i = 0; ; i ++) {
15 if(!mex[i]) {
16 return SG[n] = i;
17 }
18 }
19 }
20 int main() {
21 ios::sync_with_stdio(false);
22 cin.tie(0);
23 cout.tie(0);
24 while(cin >> k, k) {
25 for(int i = 0; i < k; i ++) cin >> f[i];
26 sort(f,f+k);
27 memset(SG, -1, sizeof(SG));
28 SG[0] = 0;
29 cin >> m;
30 while(m--){
31 cin >> l;
32 y = 0;
33 for(int j = 0; j < l; j ++) {
34 cin >> x;
35 y ^= getSG(x);
36 }
37 if(y) printf("W");
38 else printf("L");
39 }
40 printf("\n");
41 }
42 return 0;
43 }
hdu 1536 S-Nim
标签：after his set bee always ogr false bsp cout
原文地址：https://www.cnblogs.com/xingkongyihao/p/9080137.html
