kiki‘s game |
| Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others) |
| Total Submission(s): 253 Accepted Submission(s): 41 |
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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? |
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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. |
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Output If kiki wins the game printf "Wonderful!", else "What a pity!". |
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Sample Input 5 3 5 4 6 6 0 0 |
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Sample Output What a pity! Wonderful! Wonderful! |
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Author 月野兔 |
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Source HDU 2007-11 Programming Contest |
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Recommend 威士忌 |
/*
题意：就是一个n*m的棋盘，棋子在右上角(1,m)，轮流走棋子，向左，向下，向左下，如果有人无法再走了，那么这个人就输了。
初步思路：递推，左下角是必败点，然后推，如果n*m是奇数那么先走的人一定输，反之则赢
#错误：直接判断奇偶会爆内存？！！要单独判断n m......单独判断还是炸
看了原题，这个内存弄错了，这不是摆明了要用java么
*/
#include<stdio.h>
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
if((n&1)&&(m&1)) printf("What a pity!\n");
else printf("Wonderful!\n");
}
}
java版的
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// write your code here
Scanner cin=new Scanner(System.in);
int n,m;
while(cin.hasNext()){
n=cin.nextInt();
m=cin.nextInt();
if(n==0&&m==0) break;
if(n%2==1&&m%2==1) System.out.println("What a pity!");
else System.out.println("Wonderful!");
}
}
}
kiki's game
时间： 2024-10-24 07:44:12