【博弈】hdu 2147 kiki’s game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)

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

题意

博弈,从左上角开始走,每次可以向左,左下,下方移动一格,且最后不能再走的人输,问kiki先手的情况下是否会赢得这次游戏。

思路

观察后不难发现,所有的必败点(前一个人必胜)的位置均为行列都为基数的点,所以只用判断起始点是否在必败点即可。

AC代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <stack>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef long double ld;
#define mp make_pair
const int maxn = 1e5 + 7;
int (void)
{
int a, b;
while (cin >> a >> b)
{
if (a == 0 && b == 0)
break;
if (a % 2 == 1 && b % 2 == 1)
{
cout << "What a pity!" << endl;
}
else
cout << "Wonderful!" << endl;
}

return 0;
}