要求:如果一个整数各位数字之和可以被9整除，那么该数就能被9整除。编写一个脚本，提示用户输入一
个整数，然后输出该整数，并告知能够被9整数。
这里只实现了四位数的判断:
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#!/bin/bash#chapter_8-8#trap 'echo "Before executing the line:$LINENO,num=$num,i=$i,sum=$sum" ' DEBUGecho "Please input a integer number:"
read num
for ((i=3;i>=0;i--))
do let divisor=10**i
let temp=num/divisor
if [ $temp -ne 0 ]
then
break
fi
doneecho this is a $((i+1)) bit number.
sum=0
case "$i" in
3) let sum=num/10**3+num%10**3/10**2+num%10**2/10+num%10 ;;
2) let sum=num/10**2+num%10**2/10+num%10 ;;
1) let sum=num/10+num%10 ;;
0) let sum=num ;;
esacecho sum=$sum
if [ $((sum%9)) -eq 0 ]
then echo "$num can be divided by 9!"
else echo "$num can not be divided by 9!"
fi |
执行脚本效果:
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[root@localhost charpter8]# sh 8-8
Please input a integer number:4321this is a 4 bit number.sum=10
4321 can not be divided by 9![root@localhost charpter8]# sh 8-8
Please input a integer number:432this is a 3 bit number.sum=9
432 can be divided by 9![root@localhost charpter8]# sh 8-8
Please input a integer number:9333this is a 4 bit number.sum=18
9333 can be divided by 9![root@localhost charpter8]#
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这能实现不大于4位数的数值。
谁有好的脚本思路可以共享下。
本文转自 marbury 51CTO博客，原文链接:http://blog.51cto.com/magic3/1428603