hdu 5708 Alice and Bob(尼姆博弈)

题目链接：hdu 5708

Alice and Bob

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)

Problem Description
Alice and Bob are playing a stone game in a board of n×m cells.

In the begining, the stone is in the upperleft cell. And in each turn, they can move the stone one cell to the right or one cell down, or diagonally k cells down to the right, which means if you are at (x,y), then you could move into (x+1,y), (x,y+1) or (x+k,y+k) at the next step. The player who can not move loses. They play in turns and Alice moves first.

Now given n, m and k, could you tell me who is the winner?

Input
First line contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, the first line is two integers Q and k.
In the following Q lines, each line contains n and m.(1≤Q≤1000,1≤k,n,m≤109)

Output
For each test case, output Q lines.
If Alice is the winner, output “Alice”. Otherwise “Bob”.

Sample Input
2
2 1
4 5
3 4
2 3
4 5
5 6

Sample Output
Alice
Alice
Alice
Bob

翻以前的比赛翻到这道题，之前女生赛期间对这道题根本无从下手，现在一看觉得可以做，于是画起了表，不过果然光凭画表还是比较难找规律，最后还是代码打表才找出了规律。

PS:这题过的人不多，是不会的人不会做，会的都懒得做嘛Orz。
以防万一说一下：如果有人不知道sg函数是啥就过来看这篇博客的话，你肯定连打表代码都看不懂，快去搜下sg函数和尼姆博弈。不过搜完看完，这道题基本就会了=。=

WA了一下，发现k = 1的时候还要特判一下Orz。

打表码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define M 21
#define mod 1000000007
#define INF -0x3f3f3f3f

using namespace std;

int g[M][M];

void sg(int k)
{
    for(int i = 2; i <= M; i += 2)
    {
        g[i + 1][1] = g[1][i + 1] = 0;
        g[i][1] = g[1][i] = 1;
    }

    for(int i = 2; i <= M; i++)
    {
        for(int j = 2; j <= M; j++)
        {
            if((!g[i - 1][j]) || (!g[i][j - 1]) || ((!g[i - k][j - k]) && (i - k > 0) && (j - k > 0)))
            {
                g[i][j] = 1;
                continue;
            }
            else
            {
                g[i][j] = 0;
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int k;
        scanf("%d", &k);
        sg(k);
        for(int j = 1; j <= M; j++)
        {
            for(int k = 1; k <= M; k++)
            {
                printf("%d ", g[j][k]);
            }printf("\n");
        }printf("\n\n\n\n");
    return 0;
}

ac码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define M 21
#define mod 1000000007
#define INF -0x3f3f3f3f

using namespace std;

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int q, k;
        scanf("%d %d", &q, &k);
        while(q--)
        {
            int n, m, t;
            scanf("%d %d", &n, &m);
            if(n < m)//由于表是关于主对角线对称的，为了判断方便让m保持是n和m中比较小的那个值。
            {
                t = n;
                n = m;
                m = t;
            }
            if(k == 1)
            {
                if(m % 2)
                {
                    if((n - m) % 2)
                        printf("Alice\n");
                    else
                        printf("Bob\n");
                }
                else
                {
                    printf("Alice\n");
                }
            }
            else
            {
                t = 2 * (k + 1);//循环周期
                if((m % t >= 1) && (m % t <= k))
                {
                    if((n - m) % 2)
                        printf("Alice\n");
                    else
                        printf("Bob\n");
                }
                else if(!(m % (k + 1)))
                    printf("Alice\n");
                else
                {
                    if((n - m) % 2)
                        printf("Bob\n");
                    else
                        printf("Alice\n");
                }
            }
        }
    }
    return 0;
}