kiki's game

Posted 2020-08-30 勿忘初心0924

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

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 HDU 2007-11 Programming Contest

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!");
        }
    }
}