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
LWW WWL
這道題的題目挺長的，大致就說，給定一個s集合，每次 行走的步數屬於s集合，，，共有n堆石子，問首先出手的能否必勝
話說看了好久的SG函數，，有一丁點兒領悟~~不敢誤導大家~就直接上代碼了
代碼：
/* S-Nim Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4975 Accepted Submission(s): 2141 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 */ #include#include #include #include #define LEN 110 #define MAX 10010 using namespace std ; int s[LEN] , p ,sg[MAX]; void getSG(int k) { bool hash[MAX] ; memset(sg,0,sizeof(sg)) ; for(int i = 0 ; i < MAX ; ++i) { memset(hash,false,sizeof(hash)) ; for(int j = 0 ; j < k ; ++j) { if(i-s[j]>=0) { hash[sg[i-s[j]]] = true ; } } for(int j = 0 ; j < MAX ; ++j) { if(!hash[j]) { sg[i] = j ; break; } } } } int main() { int k; while(~scanf("%d",&k) && k) { for(int i = 0 ; i < k ; ++i) { scanf("%d",&s[i]) ; } getSG(k); int m ; scanf("%d",&m) ; string ans ; while(m--) { int temp = 0 , l; scanf("%d",&l) ; for(int i = 0 ; i < l ; ++i) { scanf("%d",&p) ; temp = temp^sg[p] ; } if(temp == 0) { ans += "L"; } else { ans += "W"; } } cout<
