博弈论入门题 kiki's game

Posted 2020-10-06 joeylee97

Problem Description

Recently kiki has nothing to do. While she is bored, an idea appears in his mind, she just playes the checkerboard game.The size of the chesserboard is n*m.First of all, a coin is placed in the top right corner(1,m). Each time one people can move the coin into the left, the underneath or the left-underneath blank space.The person who can‘t make a move will lose the game. kiki plays it with ZZ.The game always starts with kiki. If both play perfectly, who will win the game?

 

 

Input

Input contains multiple test cases. Each line contains two integer n, m (0<n,m<=2000). The input is terminated when n=0 and m=0.

 

 

Output

If kiki wins the game printf "Wonderful!", else "What a pity!".

 

 

Sample Input

5 3 5 4 6 6 0 0

 

 

Sample Output

What a pity! Wonderful! Wonderful!

 

 

Author

月野兔

 

 

Source

HDU 2007-11 Programming Contest

 

 

Recommend

威士忌   |   We have carefully selected several similar problems for you:  1846 1847 1848 1849 1517 

 

 

 

显然（1,1）是必败态，因为没有操作可以选择

那么由必胜态的定义（一步之内能走到必败态的状态叫必胜态）(1,2),(2,2)，（2,1）都是必胜态

依次类推可以打一个表出来（必败态：下一步无论怎样转移都会转移到必胜态）

找规律

n,m都是奇数的时候会输！

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<functional>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

#define N 100
#define MAXN 20000 + 9
#define INF 1000000009
#define eps 0.00000001
#define sf(a) scanf("%d",&a)

int n;
int a[N];
int main()
{
    while (sf(n), n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sf(a[i]), sum ^= a[i];
        if (sum == 0)
            cout << 0 << endl;
        else
        {
            int ans = 0;
            for (int i = 0; i < n; i++)
            {
                if ((sum^a[i]) < a[i])
                    ans++;
            }
            cout << ans << endl;
        }
    }
}