, select 1
24.F2 Next
25. Select the disk, continue F2 and use the entire disk to continue F2
26. Set the name, use the TAB key to choose to actively configure the network by F2
27. Choose Asia to continue F2
28. Choose China to continue F2
29. Select PRC to continue F2
30. Use the default time to continue F2
31. Enter the root password a
7. Passing of script parameters7.1.shift commandBrief introduction:Shift n Shifts the parameter position to the left n bits at a timeExample#!/bin/bashusage () { echo "usage: ' basename ' filenames"}totalline=0# The following statement $# the number of arguments if [$#-lt 2]then usagefiwhile [$#-ne 0]do line= ' cat $ | wc-l ' echo ' $1:${line} ' totalline=$ [$totalline + $line]shiftdoneecho '----- ----"echo" Total:${totalline} "7.2.getoptsBrief introduction:Get multiple command-line arg
parameter.
Call to function
The basic format of the function call statement is:
The
The parameters in the
Example:
1 intdataarray[ -];2 intN;3 DoubleAve;4 5 //Custom Functions6 DoubleAverageintValue[],intnum) {7 intI, sum =0;8 for(i =0; i ){9Sum + =Value[i];Ten returnsum*1.0/num; One } A } - - //function Call theAve = Average (DataArray, n);Application examples of custom functions
Output multiplication Ta
Title DescriptionFor the sum of the following three digits, the sum of 2 decimal 1~a and the reciprocal of the square and 1~c of the 1~b are reserved.InputA b COutput1+2+...+a + 1^2+2^2+...+b^2 + 1/1+1/2+...+1/cSample input100 50 10Sample output47977.93TipsNote the output format, do not output extra space!Source#includemain(){int i,a,b,c;float s=0;scanf("%d%d%d",a,b,c);for(i=1;is+=i;for(i=1;is+=i*i;for(i=1;is+=1.0/i;printf("%.2f\n",s);}1016:c Language Progra
Problem DescriptionThis day the teacher gave small Hao a very simple topic relaxed: Enter a score, let you find their simplest score. InputThe first line includes a T, which represents the number of groups of test data.The next line of T lines includes a fraction. (The numerator denominator is within the int range)OutputFor each test sample, the output line has its simplest fraction.Sample Input32/63/111/36Sample Output1/33/111/36My Code:1#include 2ProcessintAintb/*greatest common divisor, so th
Problem DescriptionBeg sn=2+22+222+...+22 ... A value of 222 (with N 2). For example: 2+22+222+2222+22222 (n=5), n is entered by the keyboard.InputNOutputAndSample Input5Sample Output24690My Code:1#include 2 voidMain ()3 {4 intN,a=2, t,sum=0, I;5scanf"%d",n);6t=A;7 for(i=1; i)8 {9sum=sum+T;Tent=t*Ten+A; One } Aprintf"%d", sum); -}Other code:1#include 2 3 using namespacestd;4 5 intMain ()6 {7 Longsum=0;8 intt=2;9 intN;TenCin>>N; One for(intI=0; ii) A { -Sum + =T;
Problem DescriptionThe teacher gave a title to Xiao Hao: give you two integers x and n ( -10now that little Howe is in a quandary, please help him solve the problem. InputEnter two integers x and n per line.Outputfor each set of test data, the output line evaluates. Sample Input1 12 2Sample Output16HINT1#include 2 3 4 Long LongSumintXintN)5 {6 Long Longs=0;7 inti,j;8 Long LongS1;9 for(i=1; i)Ten { Ones1=1; A for(j=1; j) - { -s1*=x; the } -s+=S1; - } -
Problem DescriptionThe output Yang Hui the first n rows of the triangle.InputEnter a number n (n OutputThe output Yang Hui the first n rows of the triangle. (Note that there is no extra space at the end of the line, the number is output in%3d format)Sample Input34Sample Output 1 1 1 1 2 1 1 1 1 1 2 1 1 3 3 1HINTNote that there are multiple sets of inputs. Output a blank line after each set of test data.while (scanf ("%d", n)! = EOF){......}1#include 2 i
Problem Descriptionn individuals in a circle, numbered from 1 to n sequentially. From the person numbered 1 starts 1 to K, where the number of people who count K out of the circle, the output of the last left a person's original numbers.InputFirst enter a T, which indicates that there is a T group of data (1Then there are t-lines, each with 2 positive integers n and K. (1OutputFor each set of test data, output a number that represents the number of the last person left.Sample Input310 37 15 4Sam
Problem DescriptionEnter n (nInputExample of multi-group test sample. The first row of each group has an integer n indicating that there are n strings. Next there are n rows, one string per line.OutputOutput a sequential string, with each line outputting a string.Sample Input3abaaabcabSample OutputAababacab1#include 2#include string.h>3 voidSortChar* str[],intsize)4 {5 inti,j;6 Char*tmp;7 for(i=0; i1; i++)8 {9 for(j=i+1; j)Ten { One if(strcmp (str[i]
Problem DescriptionEnter n, and the output corresponds to a hollow positive hexagon with a side length of N.For easy viewing, the sample midpoint '. ' Represents a space and prints a space instead of a small dot when plotting a drawing.InputEdge length N. (NOutputSide length is a positive hexagon of nSample Input5Sample Output.....*****....*.....*...*.......*.. *.........*.*...........*.. *.........*...*.......*....*.....*.....*****1#include 2#include 3 voidPrtCharCintcount)4 { 5 while(cou
Problem DescriptionEnter n strings (nInputMultiple sets of test data. The first row of each set of test data contains an integer n, representing a total of n strings. Next, each line contains a string that consists of printable characters.OutputEach set of test sample output one row. The output finds a string that satisfies the test instructions.Sample Input3djdlkfjsadfjwedlkfjdlkfjl;jf;sfjdsl;al/dljfd2dlkasfjmmlld;femflsad;fiwejdifSample OutputDlkfjdlkfjl;jf;sfjdsl;al/ld;femflsad;fiwejdif1#incl
Problem DescriptionVerify Goldbach conjecture that any even number of sufficiently large (>=4) can be represented by a sum of two primes.InputEnter an even n. (2OutputFind A, B make n=a+bWhere A and B are two primes, and aSample Input4100Sample Output2 23 971#include 2 3#include 4 5 using namespacestd;6 7 intPrimeintm)8 9 {Ten One intI,n; A - if(m==1)return 0; - theN= (int) sqrt (Double) m); - - for(i=2; i) - + if(m%i==0)return 0; - +
The title description has n integers, so that the previous number in the order to move backward m position, the last m number into the front number of m, see figure. Write a function: To achieve the above function, in the main function input n number and output adjusted n number. Enter the number of input data n n integer moved position m output moving N number sample input101 2 3 4 5 6 7 8 9 102Sample output1#include 2 #defineN 1003 4 //Move Once5MoveintA[],intN)6 {7 intI, temp = a[n-1];8
The title describes the value of sn=1!+2!+3!+4!+5!+...+n!, where n is a number. Input n output and sample input5Sample output1531#include"stdio.h"2 3 intMainintargcChar Const*argv[])4 {5 6 inti, N;7 Long Long intSn =0, factor =1;8scanf"%d", n);9 Ten for(i =1; I ) One { ASn + =factor; -Factor = factor * (i +1); - } the -printf"%lld\n", Sn); - return 0; -}1014 C Language Programming Tut
The title describes the print out of all "daffodils", the so-called "Narcissus number" refers to a three-digit number, the number of its members of the cubic and equal to that of itself. For example: 153 is a narcissus number, because 153=1^3+5^3+3^3. Output:Input no output All narcissus number, from small start. One per line1#include 2 3 intMainintargcChar Const*argv[])4 {5 inti;6 for(i = -; I +; i++)7 {8 intGE = i%Ten;9 intBai = I/ -;Ten intShi = (I/Ten) %
Title Description Enter a line of characters, respectively, the number of letters, numbers, spaces and other characters. Input line character output statistic sample inputaklsjflj123 sadf918u324 asdf91u32oasdf/. '; 123Sample output23 16 2 41#include"stdio.h"2 3 intMainintargcChar Const*argv[])4 {5 Chars[Bayi];6 intI, Char_count =0, Num_count =0, Space_count =0, Other_count =0;7 //scanf ("%s", s);8 gets (s);9 Ten for(i =0; S[i]! =' /'; i++) One { A if(S[i] >='A' S[i]
The title description writes a function, entering a four-digit number, which requires the output of these four numeric characters, but spaces between each of the two digits. If you enter 1990, you should output "1 9 9 0". Enter a four-digit output to increase the space output sample input1990Sample output1#include 2 3 voidFuncintN)4 {5 intGE = nTen;6 intQian = n/ +;7 intBai = (N-qian * +) / -; 8 intShi = (N-qian * +-Bai * -) /Ten; 9 Tenprintf"%d %d%d%d\n", Qian, Bai, Shi, ge); On
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