Pattern - Star Pyramid
C program to print
following pyramid pattern of stars n=5
*
* * *
* * * * *
* * * * * * *
* * * * * *
* * *
Input Format
Input contains n
Constraints
1 <= n <= 20
Output Format
Print the pattern
Sample Input 0
5
Sample Output 0
*
* * *
* *
* * *
* * *
* * * *
* * * * * * *
* *
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int row, c, n, s;
scanf("%d", &n);
s = n;
for (row = 1; row <= n;
row++) // Loop to print rows
{
for (c = 1; c < s;
c++) // Loop to print spaces in a row
printf(" ");
s--;
for (c = 1; c <= 2*row -
1; c++) // Loop to print stars in a row
printf("* ");
printf("\n");
}
return 0;
}
Compete
Cell
There is
a colony of 8 cells arranged in a straight line where each day every cell
competes with its adjacent cells(neighbour). Each day, for each cell, if its
neighbours are both active or both inactive, the cell becomes inactive the next
day, otherwise it becomes active the next day.
Assumptions:
The two cells on the ends have single adjacent cell, so the other adjacent cell
can be assumed to be always inactive. Even after updating the cell state.
consider its previous state for updating the state of other cells. Update the
cell information of all cells simultaneously. Write a function cellCompete
which takes takes one 8 element array of integers cells representing the
current state of 8 cells and one integer days representing the number of days
to simulate. An integer value of 1 represents an active cell and value of 0
represents an inactive cell.
Input Format
Input
will have 8 array values and the no of days
Constraints
array
size is 8 integers
Output Format
print the
array
Sample Input 0
1 0 0 0 0 1 0 0
1
Sample Output 0
0 1 0 0 1 0 1 0
#include <stdio.h>
void cellCompete(int[],int);
int main() {
int arr[10],i,n;
for(i=0;i<8;i++)
scanf("%d
",&arr[i]);
scanf("%d",&n);
cellCompete(arr,n);
for(i=0;i<8;i++)
printf("%d
",arr[i]);
return 0;
}
void cellCompete(int arr[],int n){
int arr1[10],i;
for(;n>=1;n--)
{
arr[-1]=0;
arr[8]=0;
for(i=0;i<=7;i++)
arr1[i]=arr[i-1]^arr[i+1];
/*copying arr1 to arr */
for(i=0;i<=7;i++)
arr[i]=arr1[i];
}
}
Count the
occurrence of substring
Find the
occurrence of a sub string in a parent string
Input Format
Input
contains the string and the sub string
Constraints
1<=substring_length
<= string_length<=1000
Output Format
print the
count
Sample Input 0
hgjghjhab
ab
Sample Output 0
1
#include <stdio.h>
#include <string.h>
int main() {
int i,j,k,ctr=0;
int l1,l2;
char str[1000],fstr[1000];
fgets(str,1000,stdin);
fgets(fstr,1000,stdin);
l1=strlen(str);
l2=strlen(fstr);
for(i=0;str[i]!='\0';i++)
{
for(j=0,k=i;fstr[j]!='\0'
&& str[k]!='\0' ;j++,k++)
{
if(str[k]!=fstr[j])break;
}
if(j==l2){
ctr++;
i=k-1;
}
}
printf("%d",ctr);
return 0;
}
Eliminate
Repeated Elements
Given two
positive integer arrays ary1 and arr2 of lengths len1 and len2 respectively.
write a program to count the number of elements which are not common In the
arrays.
The input
to the function distinctElementCount of two arrays arr1 and arr2 and their
lengths len1 and len2 respectively.
The
function return the number of elements which are not common in both arrays.
Example.
arr1 =
{1, 2,3, 4, 5, 6, 7, 8, 9, 10}, lent = 10
arr2 =
{11, 12, 13, 4, 5, 6, 7, 18, 19, 20}, len2 = 10
The
distinct elements are 1, 2, 3, 8, 9, 10, 11, 12, 13, 18, 19 and 20 so the
function should return 12.
Input Format
Input
contains the length of the arrays and the values
Constraints
1 ≤ n
≤105
1 ≤
values ≤ 109
Output Format
Print the
count
Sample Input 0
5 6
34 89 12 45 93
12 93 45 23 78 35
Sample Output 0
5
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int distinctElementCount(int arr1[],int arr2[],int len1,int len2)
{
int ctr=0;
int i,j;
for(i=0;i<len1;i++)
{
for(j=0;j<len2;j++)
{
if(arr1[i]!=arr2[j])
continue;
else
break;
}
if(j==len2)
ctr++;
}
for(i=0;i<len2;i++)
{
for(j=0;j<len1;j++)
{
if(arr2[i]!=arr1[j])
continue;
else
break;
}
if(j==len1)
ctr++;
}
return ctr;
}
int main() {
int a1[105],a2[105];
int i,n1,n2;
int ctr;
scanf("%d%d",&n1,&n2);
if(n1>=1 &&
n1<=105)
if(n2>=1 &&
n2<=105)
{
for(i=0;i<n1;i++) scanf("%d",&a1[i]);
for(i=0;i<n2;i++)scanf("%d",&a2[i]); ctr=distinctElementCount(a1,a2,n1,n2);
printf("%d",ctr);
}
/* Enter your code here. Read
input from STDIN. Print output to STDOUT */
return 0;
}
GCD Of N
Numbers
Write a
'C' program to find the GCD of N numbers
Input Format
Input
will have the no of elements and the values
Constraints
1 ≤ noe ≤
100000
1 ≤
values ≤ 100000000007
Output Format
print the
GCD
Sample Input 0
5 2 4 8 16 32
Sample Output 0
2
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to find gcd of array of
// numbers
int findGCD(int arr[], int n)
{ int i;
int result = arr[0];
for (int i = 1; i < n;
i++)
result = gcd(arr[i],
result);
return result;
}
int main() {
int a[100],n,i;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
int res=findGCD( a,n);
printf("%d",res);
return 0;
}
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