kiki's game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others) Total Submission(s): 4972 Accepted Submission(s): 2908
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
Recommend
威士忌 题意：
在一个n*m的棋盘上，从 （1,m），即右上角开始向左下角走。
下棋者只能往左边(left)，左下面(left-underneath)，下面(underneath)，这三个方格下棋。
最后不能移动的人算输 思路： 手动可以画出必胜态以及必败态的图 可以很容易 找出规律 (1) 所有终结点是必败点（P点）；(2)从任何必胜点（N点）操作，至少有一种方法可以进入必败点（P点）；
(3)无论如何操作，从必败点（P点）都只能进入必胜点（N点）.
由此可知 10*10之内的为 （完全可以手写出来）
NNNNNNNNNN
PNPNPNPNPN NNNNNNNNNN PNPNPNPNPN NNNNNNNNNN PNPNPNPNPN NNNNNNNNNN PNPNPNPNPN NNNNNNNNNN PNPNPNPNPN可以看出 NN
PN 这样的规律 之后很容易知道怎么做了
#include还可以用SG函数打表找出我们所要的结果NP 图int main(){ int n,m; while(scanf("%d%d",&n,&m)) { if(n==0&&m==0)break; if(n%2==1&&m%2==1)printf("What a pity!\n"); else printf("Wonderful!\n"); } return 0;}
#include#include const int sz=200;int SG[sz][sz];int dir[3][2]={-1,0,0,1,-1,1};int n,m;void get_sg(){ int i,j; memset(SG,0,sizeof(SG)); for(i=n;i>=1;i--) for(j=1;j<=m;j++) { if(!SG[i][j]) for(int k=0;k<3;k++) { int xx=i+dir[k][0]; int yy=j+dir[k][1]; if(xx>=1&&xx<=n&&yy>=1&&yy<=m) { SG[xx][yy]=1; } } }}int main(){ int i,j; while(scanf("%d %d",&n,&m)!=EOF) { get_sg(); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) if(SG[i][j]==1) printf("N"); else printf("P"); printf("\n"); } if(SG[1][m]==1) { printf("Wonderful!\n"); } else printf("What a pity!\n"); } return 0;}