HDU 1536

煙雨默嘫 — Sat, 17 Mar 2012 06:46:00 GMT

S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1782 Accepted Submission(s): 802

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

2 5 12

3 2 4 7

4 2 3 7 12

5 1 2 3 4 5

2 5 12

3 2 4 7

4 2 3 7 12

Sample Output

LWW

WWL

//很郁闷，因为题目看不懂，只能看别人的代码。才知道意思……

//题意：有一个集合有S个数。玩N次取石子游戏，每次只能取S集合里的数个石子。求判先手胜负。

1 //递归写的
2 #include
3 using namespace std;
4
5 int t;
6 int num[101];
7 int sg[10001];
8
9 int Get_sg(int n)
10 {
11     if(n==0) return 0;
12     int i;
13     bool mex[101]={0};
14     for(i=0;i15     {
16         if(n>=num[i])
17         {
18             int temp=n-num[i];
19             if(sg[temp]==-1)
20             {
21                 sg[temp]=Get_sg(temp);
22             }
23             mex[sg[temp]]=true;
24         }
25     }
26     for(i=0;;i++)
27     {
28         if(!mex[i])
29         {
30             return i;
31         }
32     }
33 }
34
35 int main()
36 {
37     int i,l,m,k;
38     while(scanf("%d",&t)!=EOF,t)
39     {
40         for(i=0;i41         {
42             scanf("%d",&num[i]);
43         }
44         scanf("%d",&l);
45         memset(sg,-1,sizeof(sg));
46         while(l--)
47         {
48             int temp=0;
49             scanf("%d",&k);
50             while(k--)
51             {
52                 scanf("%d",&m);
53                 if(sg[m]==-1)
54                 {
55                     sg[m]=Get_sg(m);
56                 }
57                 temp^=sg[m];
58             }
59              if(temp)
60             {
61                 printf("W");
62             }
63             else
64             {
65                 printf("L");
66             }
67         }
68         putchar('\n');
69     }
70     return 0;
71 }
72

煙雨默嘫 2012-03-17 14:46 发表评论

HDU 2074叠筐

煙雨默嘫 — Sat, 17 Mar 2012 06:38:00 GMT

叠筐

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

需要的时候，就把一个个大小差一圈的筐叠上去，使得从上往下看时，边筐花色交错。这个工作现在要让计算机来完成，得看你的了。

Input

输入是一个个的三元组，分别是，外筐尺寸n（n为满足0

Output

输出叠在一起的筐图案，中心花色与外筐花色字符从内层起交错相叠，多筐相叠时，最外筐的角总是被打磨掉。叠筐与叠筐之间应有一行间隔。

Sample Input

11 B A

5 @ W

Sample Output

AAAAAAAAA

ABBBBBBBBBA

ABAAAAAAABA

ABABBBBBABA

ABABAAABABA

ABABABABABA

ABABAAABABA

ABABBBBBABA

ABAAAAAAABA

ABBBBBBBBBA

AAAAAAAAA

@@@

@WWW@

@W@W@

@WWW@

@@@

1 #include<iostream>

2 using namespace std;
3
4 char map[85][85];
5
6 int main()
7 {
8     int n,i,j,mark;
9     char a,b,t=0;
10     while(scanf("%d %c %c",&n,&a,&b)!=EOF)
11     {
12         if(t) putchar('\n');
13         if((n-1)%4)
14         {
15             t=a;
16             a=b;
17             b=t;
18         }
19         t=a,mark=1;
20         for(i=0;i<=n/2;i++)
21         {
22             t=a;
23             mark=1;
24             for(j=0;j25             {
26                 if(n!=1&&i==0&&(j==0||j==n-1))//注意n==1的时候
27                 {
28                     putchar(' ');
29                     map[i][j]=' ';
30                     continue;
31                 }
32                 map[i][j]=t;
33                 putchar(t);
34                 if(i>j||j>=(n-1)-i)//字符变换的范围
35                 {
36                     if(mark)
37                     {
38                         t=b;
39                         mark=0;
40                     }
41                     else
42                     {
43                         t=a;
44                         mark=1;
45                     }
46                 }
47             }
48             putchar('\n');
49         }
50         for(i=n/2-1;i>=0;i--)
51         {
52             for(j=0;j53             {
54                 printf("%c",map[i][j]);
55             }
56             putchar('\n');
57         }
58     }
59     return 0;
60 }

煙雨默嘫 2012-03-17 14:38 发表评论

C++博客-煙雨默嘫-随笔分类-HDU

HDU 1536

HDU 2074叠筐