can anybody tell me the concept to print pattern $ $ $ $ $ $ $?
816710
Create an outer for loop that will loop from 7 down to 0, with a step of 2.\n\nCreate one inner loops.\n\nThe inner loop should take the value of i, place it in another variable and count down to 0.\nThen print out the dollar signs for each iteration of this loop.
240615
I am giving the solution in C++ using recursion..\n\n!!\n\n#include<iostream.h>\nvoid printstars(int n)\n{\nfor(int k=0;k<n;k++)\n     cout<<"*";\ncout<<endl;\n}\nvoid print1(int n)\n{\nif(n<=1)\n   printstars(n);\nelse\n   printstars(n-2);\n}\nvoid print2(int m,int n)\n{\nif(m==n)\n   cout<<endl;\nelse \n{\n   printstars(m);\n   print2(m+2,n);\n}\n}\n\nmain()\n{\nint n,m=1;\ncout<<"Enter an odd number n";\ncin>>n;\nprint2(m,n);\nprint1(n);\n}\n\nThis produces the output u needed and also uses recursion.
803702
Check out this web page: http://mathworld.wolfram.com/Net.html
234969
Print a paper that looks like this.\n\n\nDate:______________\n\nStart Time:______________\nEnd Time:_______________\n\nDuration:____________
604105
It's called ASCI art. There's a software that can produce them using software called bg_asci
101541
This is kinda wierd without drawing it out, but I'll try to explain and you can come up with the picture.\n\n1)  Start Program\n2)  Accept Input\n3)  Add 1 to counter\n4)  If counter is not equal to 3; go to #2\n5)  If N2 > N3, do #6-8; if not, go to #9\n6)  Move N3 to a hold field\n7)  Move N2 to N3\n8)  Move the hold field to N2\n9)  if N1 > N2, do #6-8 with the N1 & N2 replacing each other\n10) if N2 > N3, do #6-8\n11) print them
860477
Here is a program which first computes all the values and then, print it in the actual table manner. If value is 34, table for 1 to 10 will be printed first, then from 11 to 20, 21 to 30 and then from 31 to 34. \n\n#include <stdio.h>\n#include <conio.h>\n\nvoid main()\n{\n clrscr();\n int a[50][12];\n int n;\n printf("Enter the number till which you need tables ");\n scanf("%d",&n);\n for(int i= 1;i<=n;i++)\n  for(int j=1;j<=12;j++)\n   a[i][j] = i*j;\n\n for(int k=1;k<=n/10;k++)\n {\n\n  for(i = 1;i<=12;i++)\n  {\n   for(int j = (k-1)*10+1;j<=(k-1)*10+10;j++)\n    printf("%d&#92;t",a[j][i]);\n   //printf("&#92;n");\n  }\n  printf("&#92;n");\n  getch();\n }\n\n for(i = 1;i<=12;i++)\n  {\n   for(int j = (n/10)*10+1;j<=n;j++)\n    printf("%d&#92;t",a[j][i]);\n   printf("&#92;n");\n  }\n  printf("&#92;n");\n\n}
364948
# extract all "words" from a string\n  while ($string =~ /(&#92;w+)(?# match a "word" )/g) {\n    print "Matched `$1'&#92;n";\n  }\n\n----------------------------------------------------------\n\nUsing /&#92;G.../gc in scalar context allows you to "inchworm" your way through a string, as you've just seen. But a more practical application of this is extracting quoted text from a string:\n\n  # modified from code by merlyn (Randal Schwartz)\n  # on the Perl Beginner's mailing list\n  \n  $_ = q{whatever "this" 'line is'};\n  \n  push @elements, $1 while\n    /&#92;G&#92;s*"([^"]*)"/gc or\n    /&#92;G&#92;s*'([^']*)'/gc or\n    /&#92;G&#92;s*(&#92;S+)"/gc;\n  print map "<<$_>> ", @elements;\n  # <<whatever>> <<this>> <<line is>>
991890
Try the 2-2 pattern. it's the best way. The 3 is done with that criss-cross pattern, but the 2-2 allows you to continue the circle, just make it have a big radius
914868
I really can't answer that however I can offer a solution to allow you to print them all.   \n\nHigh light all the answers, in your print properties hit print selection.
226919
You are correct in how the programming is running.  However the '9' IS NOT garbage data.  The PrintF function does return the total number of characters printed if you are looking for it.  it does it internal to the printf function so you don't need a return on your sub routine.\n\nhttp://www.cplusplus.com/ref/cstdio/printf.html
1071450
Are you referring to Fibonacci knitting? If so, here are some good sites that explain the mathematical theory behind it and how to incorporate it into your knitting. Using this method can help you create cool stripe patterns in sweaters, felted bags and other knitted projects.\n\nBest of luck in your knitting adventures!\n\nIndigoMuse
185636
Star pattern.\n\n.........*.........\n...................\n*.....*.....*.....*\n.....*.......*.....\n.........*.........\n...................\n*.................*
150391
It is a strings proccessing alg which uses a failure function to improve the run time rather than bm alg\nrun time:m+n\nm:the length of the pattern\nn: the length of the text\nwe use tis algorithm to find the pat string in the text string..\nmore details \nmail me at ziko@math.com\ni can provid a pascal source for it
685370
you write the digits and then draw the bar above the repeating ones.  for example, if you wanted to write out 1 1/3 as a decimal, you would notice that it is 1.333333... so it would be written as 1.3 or (more often) 1.33, with a bar over only the "3"s.  if it is a number such as 2.345676767676... the bar would go above only the first set of "67," and not over the 3, 4, or 5, since they don't repeat.  some numbers may have only one repeating digit, and others may have many.  the bar goes over the first repetition only, and is not used in cases where there are other numbers that do not repeat interrupting the pattern
52716
I hope you can understand the pseudocode, it's pretty loose.\n\nConstruct two arrays P and Q.\nLet P be empty, and Q contain the n numbers you want to permute.\nTo start, call the function PrintPermute below passing in the arrays P and Q.\n\nPrintPermute( P, Q)\n{\n  if Q is empty, print the values in P in order and return\n  else\n\n  for each value q in Q,\n  {\n       Create the strings P' and Q' where P' = P + q and Q' = Q - q\n       Call PrintPermute(P', Q')\n }\nreturn\n}
755601
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!