Basic procedural exercises under the shell

1. Simple progress bar Program

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Lable array is equivalent to the progress bar forward, the refreshed circle, progress bar to show progress, with count% display

Program run Result: 650) this.width=650; "Src=" Http://s5.51cto.com/wyfs02/M02/82/F2/wKiom1dmmevzR36jAAAK-J8K68U715.png-wh_ 500x0-wm_3-wmp_4-s_2478296629.png "title=" PROCB. PNG "width=" 952 "height=" "border=" 0 "hspace=" 0 "vspace=" 0 "style=" width:952px;height:40px; "alt=" Wkiom1dmmevzr36jaaak-j8k68u715.png-wh_50 "/>

The summation between the 2.1-100

1> with while implementation 2>

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Program Run Result:

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3> Recursive implementation

Digui

function Add ()

{

Local val=$1

Local sum=0

Local add1=0

Local add2=0

If [$val-eq 1];then

Echo 1

Return

Fi

add1= $val

Let val--

add2=$ (add $val)

Let sum= $add 1+ $add 2

Echo $sum

}

function Main ()

{

Read n

Add $n

}

Main

Program Run Result:

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4> using SEQ and BC for summation

Seq N: Generate all numbers to this number

|BC: Use |BC to sum

For example:

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So 1-100 sums can be written like this:

650) this.width=650; "Src=" Http://s1.51cto.com/wyfs02/M02/82/F1/wKioL1dmoFyz2SPpAABaQjNVLQU285.png-wh_500x0-wm_3 -wmp_4-s_158030029.png "title=" 1-100BC. PNG "alt=" Wkiol1dmofyz2sppaabaqjnvlqu285.png-wh_50 "/>

Where tr represents substitution , seq comes out of all the numbers have newline characters, by TR will ' \ n ' replaced by ' + '. The next sed is to replace the last ' + ' with ' \ n ' and then sum it through BC.

3. To find the nth Fibonacci number

1> Non-recursive implementation

Locally modified variables are local variables

the function fib ()

87 {

Val=$1 Local

[$val-eq 0-o $val-eq 1]

Do

Echo 1

Return

Do

94 I=1

J=1

[$val-ge 2]

Do

fibn= $i + $j

I= $j

101 j= $fibn

102 ((val--))

103 Done

105}

106 function Main ()

107 {

108 Read n

109 Fib $n

Echo $fibn

112}

113 Main

2> Recursive implementation

Thought: When N=1 or n=0, return 1. Since the nth Fibonacci number is the sum of the number of n-1 and n-2, the value of the N value passed is first minus 1 in the recursive save values, and then the N minus 1 recursively save the value, Nth is the sum of two of the depositary.

The function fib ()

116 {

117 Local Val=$1

118 Local Fib1=0

119 Local Fib2=0

Fibc=0

121 If [$val-eq 0-o $val-eq 1];then

122 Echo 1

123 return

124 fi

val--

126 fib1=$ (fib $val)

127 Let val--

fib2=$ (fib $val)

129 let fibc= $fib 1+ $fib 2

Echo $FIBC

131

132}

133

134 Function Main ()

135 {

136 Read N

137 fib $n

138

139}

Main

Program Run Result:

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This article is from the "Output Diamond pattern" blog, so be sure to keep this source http://10541571.blog.51cto.com/10531571/1790879

Basic procedural exercises under the shell