Write a program to count the number of times the digit is present?

Answer 1

#include<stdio.h>

#include<conio.h>

void main( )

{

int a[50],i,n,s,t=0;/* the digit 50 can be any integer, its a user defined function a[limit]*/

clrscr( );

printf("\n\nenter the number of elements u want to enter");

scanf("%d",&n);

printf("\n\nenter the numbers u desire");

for(i=0;i<n;i++)

scanf("%d",&a[i]);

printf("enter the element to be counted");

scanf("%d",&s);

for(i=0;i<n;i++)

{

if(s==a[i])

t++;

}

printf("the number u entered has occured %d number of times",t);

getch( );

}

Answer 2

// set up our input buffer

BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

String currentLine;

boolean readAnInt = false;

// keep reading input until an integer is input

while (!readAnInt) {

currentLine = in.readLine();

try {

// try to convert from string to int

Integer.parseInt(currentLine);

// now we know we have an integer: print number of digits

System.out.println("Number of Digits: " + currentLine.length());

readAnInt = true;

} catch (final NumberFormatException ex) {

// we go here if the user didn't type in an integer

}

}

Answer 3

#include<stdio.h>

#include<conio.h>

int main()

{

char a[]={'s','h','i','k','h','a','\0'};

int i,count=0;

for(i=0;a[i]!='\0';i++);

printf("%d",i);

}