Pattern Printing coding 8
CODE NO:- 10
Print the below pattern
#LOGIC
Take a variable and assign it with value (n*(n+1))/2 then keep decrementing it in the inner loop and print the pattern.
Here n is the no. of entered rows whereas (n*(n+1))/2 is the first value of pattern.
ex: if n=4 then (n*(n+1))/2 results 10.
#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;
printf("\n enter the no.of rows:");
scanf("%d",&n);
a=(n*(n+1))/2;
for(i=0;i<n;i++)
{
for(k=0;k<n-i;k++)
{
printf("%d ",a--); // post decrementing the value
}
printf("\n");
}
return 0;
}
We got this output by using DEV C++ (IDE)
Link: For other pattern printing problems
Print the below pattern
#LOGIC
Take a variable and assign it with value (n*(n+1))/2 then keep decrementing it in the inner loop and print the pattern.
Here n is the no. of entered rows whereas (n*(n+1))/2 is the first value of pattern.
ex: if n=4 then (n*(n+1))/2 results 10.
#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;
printf("\n enter the no.of rows:");
scanf("%d",&n);
a=(n*(n+1))/2;
for(i=0;i<n;i++)
{
for(k=0;k<n-i;k++)
{
printf("%d ",a--); // post decrementing the value
}
printf("\n");
}
return 0;
}
We got this output by using DEV C++ (IDE)
Link: For other pattern printing problems
Gud....heep helping
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