0
#!/bin/bash total=0 num=0 while((num<=10)); do total=`expr $total + $num` num="$num + 1" done echo "the total is $total"
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本帖最后由 lucas1202 于 2015-8-12 16:20 编辑
#!/bin/bash
a=0
sum=0
while [ $a -lt 10 ]
do
a=$[$a+1]
sum=$[$sum+$a]
done
echo $sum
#!/bin/bash
a=0
sum=0
while [ $a -lt 10 ]
do
a=$[$a+1]
sum=$[$sum+$a]
done
echo $sum
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本帖最后由 追风者 于 2015-10-2 22:43 编辑
sum=0;i=1;while( i -lt 10 ) ;do i=`seq 1 1 10`;sum=$[$i+$sum] echo sum;done
sum=0;i=1;while( i -lt 10 ) ;do i=`seq 1 1 10`;sum=$[$i+$sum] echo sum;done
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#!/bin/bash
#use 'while' to come ture the sum of 1 to 10
#
n=1
m=0
while [ $n -le 10 ]
do
m=$[$m+$n]
n=$[$n+1]
done
echo $m
#use 'while' to come ture the sum of 1 to 10
#
n=1
m=0
while [ $n -le 10 ]
do
m=$[$m+$n]
n=$[$n+1]
done
echo $m
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#!/bin/bash
##1到10所有数字相加求和
##write by 2016-01-08
sum=0
i=1
while ((i<=10)) ;do
sum=$[$sum+$i]
i=$[$i+1]
done
echo "1到10所有数字相加为：$sum"
##1到10所有数字相加求和
##write by 2016-01-08
sum=0
i=1
while ((i<=10)) ;do
sum=$[$sum+$i]
i=$[$i+1]
done
echo "1到10所有数字相加为：$sum"
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#!/bin/sh
#1 到10 相加
i=1
tot=0
while [[ $i -lt 11 ]]
do
tot=$(($tot + $i))
i=$(($i + 1))
done
echo $tot
#1 到10 相加
i=1
tot=0
while [[ $i -lt 11 ]]
do
tot=$(($tot + $i))
i=$(($i + 1))
done
echo $tot
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- #!/bin/bash
- uplimit=10
- sum=0
- i=1
- while [ $i -le $uplimit ]
- do
- sum=$(($sum + $i))
- i=$(($i + 1))
- done
- echo $sum
- #!/bin/env python
- print sum(range(1,11))
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#!/bin/bash
# to calculate the sum of the num from 1 to 10
sum=0
i=1
while(($i<11));do
sum=$(($i+$sum))
i=$(($i+1))
done
echo "1+2+...+10=$sum"
# to calculate the sum of the num from 1 to 10
sum=0
i=1
while(($i<11));do
sum=$(($i+$sum))
i=$(($i+1))
done
echo "1+2+...+10=$sum"
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#!/bin/bash
#declare -i sum=0
#for i in {1..10}
#do
# sum=$[$sum+$i]
#done
#echo "sum="$sum
declare -i sum=0
let i=1
while [ $i -le 10 ]
do
sum=$[$sum+$i]
i=$[$i+1]
done
echo "sum="$sum
#declare -i sum=0
#for i in {1..10}
#do
# sum=$[$sum+$i]
#done
#echo "sum="$sum
declare -i sum=0
let i=1
while [ $i -le 10 ]
do
sum=$[$sum+$i]
i=$[$i+1]
done
echo "sum="$sum
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#!/bin/bash
cat <
EOF
number=1
while [ $number -le 10 ];do
#count=$(($count+$number))
count=$[$count+$number]
if [ $number -eq 10 ];then
echo "$number=$count"
else
echo -n "$number+"
fi
number=$(expr $number + 1)
done
cat <
EOF
number=1
while [ $number -le 10 ];do
#count=$(($count+$number))
count=$[$count+$number]
if [ $number -eq 10 ];then
echo "$number=$count"
else
echo -n "$number+"
fi
number=$(expr $number + 1)
done
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#!/bin/bash
#writen by caibingsen
#time:2016-02-29 21:55:44
i=1
sum=0
while :
do
if [ $i == 11 ]
then
echo $sum
exit
else
sum=$[$i+$sum]
i=$[$i+1]
fi
done
#writen by caibingsen
#time:2016-02-29 21:55:44
i=1
sum=0
while :
do
if [ $i == 11 ]
then
echo $sum
exit
else
sum=$[$i+$sum]
i=$[$i+1]
fi
done
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本帖最后由 we14578 于 2016-3-1 14:04 编辑
#!/bin/bash
i=1
sum=0
while [ $i -le 10 ]
do
sum=$[$sum+$i]
i=$[$i+1]
done
echo $sum
#!/bin/bash
i=1
sum=0
while [ $i -le 10 ]
do
sum=$[$sum+$i]
i=$[$i+1]
done
echo $sum
0
#! /bin/bash
##求1到10所有数字的和
i=11
sum=0
while [ $i -ge 1 ]
do
i=$[ $i - 1 ]
sum=$[ $sum + $i ]
done
echo $sum
最后运算的结果是一样的，就是不知道这种写法是否符合要求
##求1到10所有数字的和
i=11
sum=0
while [ $i -ge 1 ]
do
i=$[ $i - 1 ]
sum=$[ $sum + $i ]
done
echo $sum
最后运算的结果是一样的，就是不知道这种写法是否符合要求
0
本帖最后由 wsw13640218682 于 2016-4-8 22:17 编辑
- #!/bin/bash
- a=1
- sum=0
- while [ $a -le 10 ]
- do
- sum= $[sum+a]
- ((a++))
- done
- echo $sum
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