猴子喝汽水问题

 

有个猴子手里有22元，去买汽水喝，一瓶汽水是1元，三个空瓶可以换取一瓶汽水，问该猴子可以喝多少水？

 

使用shell脚本方式实现

 

 

 

vi monkey.sh

 

#!/bin/bash

read -p 'input the number of money: ' count   //钱数
read -p 'inout the average of fruit: ' bs     //单价
read -p 'input the change of fruit: ' chg     //交换价

#count=22
#bs=1
#chg=3                                        //测试使用

A=$(($count/$chg))                            //空瓶可换汽水数
B=$(($count%$chg))                            //换后的空瓶数
tmp[1]=$A
sum=0
b=$B

for i in $(seq 1  5 )
do
tmp[$(($i+1))]=$(($((${tmp[$i]}+$b))/$chg))   //计算所有空瓶可换汽水数
echo -n "tmp:" ${tmp[$(($i+1))]} " "         
b=$(($((${tmp[$i]}+$b))%$chg))                //显示中间产生空瓶数值
echo "b:"$b
sum=$(( $sum+${tmp[$(($i+1))]} ))             //计算此次可喝汽水数
done

[ $b -eq 2 ]&&sum=$(($sum+1))||sum=$sum       //如果有2个空瓶时，可借

                                              //一空瓶后喝完再还回去
echo $(($count+$A+$sum))

 

上面的是原型，下面给大家通用型的：

#!/bin/bash

read -p 'input the number of money: ' count
read -p 'inout the average of fruit: ' bs
read -p 'input the change of fruit: ' chg

A=$(($((count/bs))/$chg))
B=$(($((count/bs))%$chg))

tmp[1]=$A
sum=0
b=$B

for i in $(seq 1  9 )
do
tmp[$(($i+1))]=$(($((${tmp[$i]}+$b))/$chg))
echo -en "tmp:" ${tmp[$(($i+1))]} " "
b=$(($((${tmp[$i]}+$b))%$chg))
echo "b:"$b
sum=$(( $sum+${tmp[$(($i+1))]} ))
done

[ $b -eq $((chg-1)) ]&&sum=$(($sum+1))

echo $((count/bs))"+"$A"+"$sum"="$(($((count/bs))+$A+$sum))