kiki's game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 8946 Accepted Submission(s): 5335
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!
题意:给你一个 n*m的方阵 然后 有两个叫kiki和zz的人 kiki 每次都从 方阵的右上角开始 走 然后 zz 再开始走 zz必须在上次kiki走的空的 右方 或者 下方 或者右下方,这三个方向
谁最先走到 左下方 谁就是胜利者 .
这个大大 讲得很不错 http://blog.csdn.net/no_retreats/article/details/7900654
下面附上代码
1 #include<stdio.h> 2 int main() 3 { 4 int n,m; 5 while(scanf("%d%d",&n,&m)&&!(n==0&&m==0)) 6 { 7 if(n%2!=0&&m%2!=0) 8 printf("What a pity! "); 9 else 10 printf("Wonderful! "); 11 } 12 }