循环实操(1)求和、从1到输入整数之间所有整数的和、偶数和还有奇数和、商场购物

1、计算从1到100所有整数的和

for

#!/bin/bash
sum=0
for i in {
    
    1..100}
do
    sum=$[$i+$sum]
done
echo $sum

while

#!/bin/bash
sum=0
i=1
while [ $i -le 100 ]
do
   sum=$[$sum+$i]
   let i++
   #let i=i+1
done
echo "$sum"

2、提示用户输入一个小于100的整数,并计算从1到该数之间所有整数的和

for

#!/bin/bash
read -p "输入小于100的一个整数:" a
sum=0
for ((i=1;i<=a;i++))

do
  sum=$[$sum+$i]

done
echo "1到输入值之间的整数和为$sum"

while

#!/bin/bash
read -p "输入一个小于100的整数:" a
sum=0
while [ $a -ge 1 ]
do 

sum=$[$sum+$a]
let a--

done
echo "1到输入值之间的整数和为$sum"

3、求从1到100所有整数的偶数和、奇数和

for

#!/bin/bash
a=0
b=0
for i in {
    
    0..100..2}
do 
 a=$[$a+$i]

done
echo "奇数和为$a"


for i in {
    
    1..100..2}
do 
 b=$[$b+$i]

done
echo "偶数和为$b"

while

#!/bin/bash
a=0
b=0

i=1
c=0

while [ $i -le 100 ]
do
  a=$[$a+$i]
  let i=$[$i+2]
done
echo "奇数和为$a"


while [ $c -le 100 ]
do
  b=$[$b+$c]
  let c=$[$c+2]
done
echo "偶数和为$b"

4、写个逛淘宝选购商品脚本,每家商店有五种商品选购(衣服500元,裤子400元,鞋子350元,帽子150元,袜子50元),每次选购完或不买都会提示用户是否继续逛下一家商店,如果不再继续逛的话进行购物车结算总额。

#!/bin/bash
i=1
sum=0
while true
do
 echo "来到第$i家商店门口"
 read -p "是否进入看看(yes/no)" doing
 while [ $doing = "yes" ]
 do
  echo "1:衣服¥500"
  echo "2:裤子¥400"
  echo "3:鞋子¥350"
  echo "4:帽子¥150"
  echo "5:袜子¥50"
  echo "其他:放弃购买"
  read -p "请选择需要购买的商品序列:" num
  case $num in
  1)
   echo "衣服购买成功"
   sum=$[sum+=500]
   ;;
  2)
   echo "裤子购买成功"
   sum=$[sum+=400]
   ;;
  3)
   echo "鞋子购买成功"
   sum=$[sum+=350]
   ;;
  4)
   echo "帽子购买成功"
   sum=$[sum+=150]
   ;;
  5)
   echo "袜子购买成功"
   sum=$[sum+=50]
   ;; 
  *)
   echo "放弃购买"
            
  esac
  read -p "是否继续在这家店购物(yes/no)" doing      
 done 

 read -p "是否继续逛街(yes/no)" going
 if [ $going = "yes" ]
 then
  let i++
 else
  break
 fi 
done
echo -e "一共经过$i家店。\n合计购物总价:$sum。"

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转载自blog.csdn.net/weixin_51616026/article/details/111476072