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L1-044 稳赢

大家应该都会玩“锤子剪刀布”的游戏:两人同时给出手势,胜负规则如图所示:

img

现要求你编写一个稳赢不输的程序,根据对方的出招,给出对应的赢招。但是!为了不让对方输得太惨,你需要每隔K次就让一个平局。

输入格式:

输入首先在第一行给出正整数K(≤10),即平局间隔的次数。随后每行给出对方的一次出招:ChuiZi代表“锤子”、JianDao代表“剪刀”、Bu代表“布”。End代表输入结束,这一行不要作为出招处理。

输出格式:

对每一个输入的出招,按要求输出稳赢或平局的招式。每招占一行。

输入样例:

tex
2
ChuiZi
JianDao
Bu
JianDao
Bu
ChuiZi
ChuiZi
End
2
ChuiZi
JianDao
Bu
JianDao
Bu
ChuiZi
ChuiZi
End

输出样例:

tex
Bu
ChuiZi
Bu
ChuiZi
JianDao
ChuiZi
Bu
Bu
ChuiZi
Bu
ChuiZi
JianDao
ChuiZi
Bu

Solution:

java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        String s = in.readLine();
        int index = 0;
        StringBuilder sb = new StringBuilder();
        while (!s.equals("End")) {
            index++;
            if (index % (n + 1) == 0) {
                sb.append(s).append("\n");
            } else {
                switch (s) {
                    case "JianDao":
                        sb.append("ChuiZi\n");
                        break;
                    case "ChuiZi":
                        sb.append("Bu\n");
                        break;
                    case "Bu":
                        sb.append("JianDao\n");
                        break;
                }
            }
            s = in.readLine();
        }
        System.out.print(sb);
    }
}
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        String s = in.readLine();
        int index = 0;
        StringBuilder sb = new StringBuilder();
        while (!s.equals("End")) {
            index++;
            if (index % (n + 1) == 0) {
                sb.append(s).append("\n");
            } else {
                switch (s) {
                    case "JianDao":
                        sb.append("ChuiZi\n");
                        break;
                    case "ChuiZi":
                        sb.append("Bu\n");
                        break;
                    case "Bu":
                        sb.append("JianDao\n");
                        break;
                }
            }
            s = in.readLine();
        }
        System.out.print(sb);
    }
}

Released under the MIT License.