POJ 2960.S-Nim【博弈论】【4月5】
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S-Nim
Description
Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
recently learned an easy way to always be able to find the best move:
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.
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.
- 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.
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.
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.
Print a newline after each test case.
Sample Input
2 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120
Sample Output
LWWWWL
博弈论的问题,只把代码贴上:
#include<iostream>#include<cstring>#include<cstdio>using namespace std;int k, m, hi, num1, ans;int s[110];int vis[10100];int sg[10100];int main(){ while(scanf("%d", &k) && k) { memset(s, 0, sizeof(s)); memset(sg, 0, sizeof(sg)); for(int i = 0;i < k; ++i) scanf("%d", &s[i]); for(int i = 0;i < 10001; ++i) { memset(vis, 0, sizeof(vis)); for(int j = 0;j < k; ++j) { if(i >= s[j]) vis[sg[i - s[j]]] = 1; } for(int j = 0;j < 10001; ++j) { if(vis[j] == 0) { sg[i] = j; break; } } } scanf("%d", &m); for(int i = 0;i < m; ++i) { ans = 0; scanf("%d", &num1); while(num1--) { scanf("%d", &hi); ans ^= sg[hi]; } if(ans == 0) cout <<"L"; else cout <<"W"; } cout << endl; } return 0;}
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