Shell脚本实现的猜数字小游戏

技术  /  管理员 发布于 7年前   141

生成的密码和用户输入可以接受重复数字。
所以相对一般规则的猜数字可能难度要大不少。

本版本规则：

A--数字对，位置也对
B--排除A的结果后，数字对，但位置不对

开始后，系统化初始化一个4位可重复数字，如“1223”。假设用户第一次输入“1234”，那么系统将提示“2A1B”，前两位数字“12”相同并且位置也相同，为“2A”。后两位数字中，用户输入的“3”与密文中“3”相同，但两者位置不同，则为“1B”，最终结果为“2A1B”。

再假设用户此时输入“1232”，那么结果则为“2A2B”，计算方法与前次一样。

代码如下：

#!/bin/bashclearechoecho "###################################################################"echo "# this is a bash-shell game write by Email:[email protected] #"echo "# the game called *digits*,and this version have repeated numbers #"echo "#  version 1.0  #"echo "###################################################################"echo -e "\n\n"declare INPUTdeclare PASSWORDdeclare Adeclare Bdeclare Xdeclare Ydeclare LOOP#This funtion init the variable PASSWORD that user need to guessinit_password(){    PASSWORD=`echo $(($RANDOM%10000))`    echo $PASSWORD | grep '^[0-9]\{4\}$' >/dev/null 2>&1    if [ $? != 0 ]    then        init_password    else        input    fi}#This funtion accept the input from user's keyboardinput(){    echo -n "please input a number between 0000-9999:"    read INPUT    echo $INPUT | grep '^[0-9]\{4\}$' >/dev/null 2>&1    if [ $? != 0 ]    then        echo "retry a number between 0000-9999 and do not input a char"        input    else        judge    fi}#This funtion is the main funtionjudge(){    X=$INPUT    Y=$PASSWORD    while [ $INPUT != $PASSWORD ]    do        A=0        B=0        judge_a        judge_b        LOOP=`expr $LOOP + 1`        echo "****************************"        echo "*      "$A"A"$B"B      *"        echo "****************************"        input    done}#This funtion count the variable A's valuejudge_a(){        for i in `seq 4`        doVAR_INPUT=`expr substr "$X" $i 1`for j in `seq 4`do    VAR_PASSWORD=`expr substr "$Y" $j 1`    if [[ $VAR_INPUT = $VAR_PASSWORD && $VAR_INPUT != "" && $VAR_PASSWORD != "" && $i = $j ]]    then        A=`expr $A + 1`        X=`expr substr $X 1 "$[$i-1]"``expr substr $X "$[$i+1]" 4`        Y=`expr substr $Y 1 "$[$i-1]"``expr substr $Y "$[$i+1]" 4`        judge_a    fidone        done}#This funtion count the variable B's valuejudge_b(){        for i in `seq 4`        doVAR_INPUT=`expr substr "$X" $i 1`for j in `seq 4`do    VAR_PASSWORD=`expr substr "$Y" $j 1`    if [[ $VAR_INPUT = $VAR_PASSWORD && $VAR_INPUT != "" && $VAR_PASSWORD != "" ]]    then        B=`expr $B + 1`        X=`expr substr "$X" 1 "$[$i-1]"``expr substr "$X" "$[$i+1]" 4`        Y=`expr substr "$Y" 1 "$[$j-1]"``expr substr "$Y" "$[$j+1]" 4`        judge_b    fidone        done}#This is the begin of scriptLOOP=1init_passwordecho "#############################################"echo "#congratulations!You have tried $LOOP times!  #"echo "#    The password is $PASSWORD !       #"echo "#############################################"