HDU1536:S-Nim(sg函数)
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S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10368 Accepted Submission(s): 4262
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1536
Descripion:
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:
Sample Output:
LWW WWL
题意:
n堆石子,每堆石子都有相应的个数,然后现在每个人选择一堆从种取石子,注意这里他一开始给出了一个集合S,取的个数是集合S里面的数。最后问先手还是后手赢。
题解:
sg函数的一个简单应用吧,最后求sg函数的时候稍微修改一下就行了。
具体见代码吧。
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <cmath> #include <vector> using namespace std; typedef long long ll; const int N = 10005; int n, m; int s[N], sg[N]; short check[N]; void Get_sg() { memset(sg, 0, sizeof(sg)); for(int i = 1; i < N; i++) { memset(check, 0, sizeof(check)); for(int j = 1; j <= n && s[j] <= i; j++) { check[sg[i - s[j]]] = 1; } for(int j = 0; j < N; j++) { if(!check[j]) { sg[i] = j; break ; } } } } int main() { while(scanf("%d", &n) && n) { for(int i = 1; i <= n; i++) scanf("%d", &s[i]); sort(s + 1, s + n + 1); Get_sg(); scanf("%d", &m); while(m--) { int k, t, x = 0; scanf("%d", &k); for(int i = 1; i <= k; i++) { scanf("%d", &t); x ^= sg[t]; } if(x) printf("W"); else printf("L"); } printf(" "); } return 0; }
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