kiki's game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 10656 Accepted Submission(s): 6455
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
在一个n*m的棋盘内,从右上角(1, m)点出发,每次可以进行的移动是:左移一,下移一,左下移一。然后kiki每次先走,判断kiki时候会赢(对方无路可走的时候)。
左下角(n, m)为必败态,把PN状态的点描绘出来,规律是n和m中至少有一个偶数时必胜,
1 #include"stdio.h" 2 3 int main( ) 4 { 5 int n,m; 6 while(scanf("%d%d",&n,&m)&&(n!=0||m!=0)) 7 { 8 if(n%2==0||m%2==0) 9 printf("Wonderful! "); 10 else 11 printf("What a pity! "); 12 } 13 return 0; 14 }