hdu 1536/ hdu 1944 S-Nim（sg函数）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7751    Accepted Submission(s): 3266

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

 

Source

 

 

题意：首先输入K 表示一个集合的大小  之后输入集合 表示对于这对石子只能去这个集合中的元素的个数

之后输入 一个m 表示接下来对于这个集合要进行m次询问

之后m行 每行输入一个n 表示有n个堆  每堆有n1个石子  问这一行所表示的状态是赢还是输 如果赢输入W否则L

思路：对于n堆石子 可以分成n个游戏 之后把n个游戏合起来就好了

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXN =  + ;
 const int MAXM =  + ;

 int f[MAXN];//f[0]存合法移动个数
 int sg[MAXM];
 bool exist[MAXN];//hash, sg不会超过合法移动个数MAXN

 void getSg(int n)
 {
     int i, j;
     sg[] = ;
     for (i = ; i <= n; ++i) {
         memset(exist, false, sizeof(exist));
         for (j = ; j <= f[] && f[j] <= i; ++j) {
             exist[sg[i - f[j]]] = true;
         }
         for (j = ; j < MAXN; ++j) {
             if (!exist[j]) {
                 sg[i] = j;
                 break;
             }
         }
     }
 }

 int main()
 {
     int k;//, s;
     int m;
     int l, hi;
     int i, j;
     int sum;

     while (~scanf("%d", &k)) {
         if (k == ) {
             break;
         }
         f[] = k;
         for (i = ; i <= k; ++i) {
             scanf("%d", &f[i]);
         }
         sort(f + , f +  + k);
         getSg();

         scanf("%d", &m);
         for (i = ; i < m; ++i) {
             scanf("%d", &l);
             sum = ;
             for (j = ; j < l; ++j) {
                 scanf("%d", &hi);
                 sum ^= sg[hi];
             }
             if (sum != ) {
                 printf("W");
             } else {
                 printf("L");
             }
         }
         printf("\n");

     }
     return ;
 }

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXN =  + ;
 const int MAXM =  + ;

 int s[MAXN];
 int sg[MAXM];
 int n;//s中的个数

 int dfsSg(int x)
 {
     if (sg[x] != -) {
         return sg[x];
     }
     int i;
     bool vis[MAXN];//sg范围
     memset(vis, false, sizeof(vis));
     for (i = ; i < n && s[i] <= x; ++i) {
         dfsSg(x - s[i]);
         vis[sg[x - s[i]]] = true;
     }
     for (i = ; i <= x; ++i) {
         if (!vis[i]) {
             sg[x] = i;
             break;
         }
     }
     return sg[x];
 }

 int main()
 {
     int k;//, s;
     int m;
     int l, hi;
     int i, j;
     int sum;

     while (~scanf("%d", &k)) {
         if (k == ) {
             break;
         }
         n = k;
         for (i = ; i < k; ++i) {
             scanf("%d", &s[i]);
         }
         sort(s, s + k);
         memset(sg, -, sizeof(sg));
         scanf("%d", &m);
         for (i = ; i < m; ++i) {
             scanf("%d", &l);
             sum = ;
             for (j = ; j < l; ++j) {
                 scanf("%d", &hi);
                 sum ^= dfsSg(hi);
             }
             if (sum != ) {
                 printf("W");
             } else {
                 printf("L");
             }
         }
         printf("\n");

     }
     return ;
 }