Pattern Printing

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Pattern Printing coding 10

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CODE NO:- 12 Print the below pattern #LOGIC Take two variables and assign it with value (((n*(n+1))/2)+2)-(n-1)   and 0(zero)  here n means the no. of entered rows and  (((n*(n+1))/2)+2)-(n-1)  is the first element of the pattern. ex: if n is 4 then   (((n*(n+1))/2)+2)-(n-1)  results 9.  Now copy the value   (((n*(n+1))/2)+2)-(n-1)  in other variable and increment it by 1 in the inner loop whereas in the outer loop decrement it by value (n-i-1). Proceed in the same manner to print the pattern. #Program( In C): In    C language we use  ' // '  for Single line comment and /*---*/ for multi line comment #include<stdio.h>// header file for i/p and o/p int main() { int n,i,k,a,b=0; printf("\n enter the no.of rows:"); scanf("%d",&n); a= (((n*(n+1))/2)+2)-(n-1) ; for(i=0;i<n;i++) {            b=a; ...
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General code 3

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CODE NO:- 18 Q) C program to convert a binary number to octal number. Input: 101 Output: 5 Input: 11010 Output: 32 Test Case: ------------- (1) Valid Input:     (a) Only number consisting of 0s and 1s will be given as input (2) Invalid inputs:     (a) Decimal ( b) Fraction  (c) String  (d) Two or more command-line arguments  (e) Negative number (3)You should generate output as follows:      (a) For right output print just the actual Octal Value to STDOUT without any other text.         (b) If any error: print 'ERROR' to the STDOUT without any other text. #LOGIC First, we need to convert the Binary no. into in Decimal no. then convert the decimal no. to octal no. . #Program( In C): In    C language we use  ' // ' for Single line comment and /*---*/ for multi line comment                             ...
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Series code 2

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CODE NO:- 06 Q)Find the no. for a user given position for this series  0,0,2,1,4,2,6,3,8,4,…     For example: (1) At position 4 you should get no. :1                             (2) At position 9 you should get no. :8 #LOGIC According to the given series, we can conclude that : 1) At even indexes, we are having a series which starts from 0 and gets incremented by (+2) . 2) At odd indexes, we are having a series which starts from 0 and gets incremented by (+1) . NOTE : Here we are considering 0 as starting index. #Program  (In C++): #include<iostream> using namespace std; int main() { int n,t; cout<<"\n enter the position :"; cin>>n; if(n%2==0) { n=n/2; t=(n-1); cout<<"The no. at given position is:"<<t; } else { n=n/2+1; t=(n-1)*2; cout<<"The...
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