Posted on 2010-08-10 20:48
MiYu 閱讀(1120)
評(píng)論(5) 編輯 收藏 引用 所屬分類:
ACM ( 組合 ) 、
ACM ( 博弈 )
MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)載自 ______________白白の屋
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2147
題目描述:
kiki's game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others)
Total Submission(s): 1806 Accepted Submission(s): 1055
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!
題目分析:
一直WA , 分析也沒(méi)分析出來(lái) , 百度了一下別人的解題報(bào)告后.............我承認(rèn)....我被征服了.....................
分析如下:
P點(diǎn)：就是P個(gè)石子的時(shí)候，對(duì)方拿可以贏(自己輸?shù)?
N點(diǎn)：就是N個(gè)石子的時(shí)候，自己拿可以贏
現(xiàn)在關(guān)于P,N的求解有三個(gè)規(guī)則
（1）：最終態(tài)都是P
（2）：按照游戲規(guī)則，到達(dá)當(dāng)前態(tài)的前態(tài)都是N的話，當(dāng)前態(tài)是P
（3）：按照游戲規(guī)則，到達(dá)當(dāng)前態(tài)的前態(tài)至少有一個(gè)P的話，當(dāng)前態(tài)是N
題意：
在一個(gè)m*n的棋盤(pán)內(nèi)，從(1,m)點(diǎn)出發(fā)，每次可以進(jìn)行的移動(dòng)是：左移一，下移一，左下移一。然后kiki每次先走，判斷kiki時(shí)候會(huì)贏（對(duì)方無(wú)路可走的時(shí)候）。
我們可以把PN狀態(tài)的點(diǎn)描繪出來(lái)：：
這些點(diǎn)的描繪有一個(gè)程序：：
【
#include<iostream>
using namespace std;
bool map[2001][2001];//1 P 0 N;
int main(){
int i,j,k;
map[1][1]=1;
for(i=2;i<=2000;i++)
{
if(map[i-1][1])
map[i][1]=0;
else map[i][1]=1;
for(j=2;j<i;j++){
if(!map[i][j-1]&&!map[i-1][j-1]&&!map[i-1][j])
map[i][j]=1;
else map[i][j]=0;
}
if(map[1][i-1])
map[1][i]=0;
else map[1][i]=1;
for(j=2;j<i;j++){
if(!map[j-1][i]&&!map[j-1][i-1]&&!map[j][i-1])
map[j][i]=1;
else map[j][i]=0;
}
if(!map[i][i-1]&&!map[i-1][i-1]&&!map[i-1][i])
map[i][i]=1;
else map[i][i]=0;
}
int M,N;
for(i=1;i<=10;i++){
for(j=1;j<=10;j++)
printf("%c ",map[i][j]?'P':'N');
printf("\n");
}
while(scanf("%d%d",&M,&N)&&M&&N){
if(map[M][N]) printf("What a pity!\n");
else printf("Wonderful!\n");
}
return 0;
}
】
具體代碼如下:
#include <iostream>
using namespace std;
int main ()
{
int n,m;
while ( cin >> n >> m , n + m )
{
puts ( n%2 && m % 2 ? "What a pity!" : "Wonderful!");
}
return 0;
}