#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( );
}
// 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
}
}
#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);
}
#include "stdio.h" int main() { unsigned int number, count; printf("Enter the Number \t"); scanf("%d", &number); printf("The even numbers are: \n"); for(count = 0x01; (count < number && number!= 0x00)) { if(count%2) { }else { printf("%d\n", count); } count++; } return 0; }
I would use a loop like this: const char *p= str-1; size_t count= 0; while (*++p) if (islower (*p)) ++count;
yyu5uty
.... String line = "This is example program with spaces"; String[] tokens = line.split(" "); System.out.println(tokens.length-1); .......
1. Find algorithm.2. Implement it.Hint: if a non-zero N number has K 1-bits, then (N AND N-1) has K-1 1-bits.
abhimanyu
Write a program to count the number of IS in any number in register B and put the count in R5.
A program which is used to count the number of numbers in an array using a 8085 microprocessor is known as a assembly language program.
#include "stdio.h" #include "conio.h" #define TABLE_UP_TO_20 20 void table_of_a_number(int number); int main() { int i = 0x00; printf("Enter a positive number in decimal whose table has to be generated"); scanf("%d",&i); table_of_a_number(i); return 0; } void table_of_a_number(int number) { int count = 0x00; for(count = 0x01;count <=TABLE_UP_TO_20 ;count++) { /* THis will print the table*/ printf(" %d * %d = %d\n", number,count, (number*count)); }
#include "stdio.h" int main() { unsigned int number, count; printf("Enter the Number \t"); scanf("%d", &number); printf("The even numbers are: \n"); for(count = 0x01; (count < number && number!= 0x00)) { if(count%2) { }else { printf("%d\n", count); } count++; } return 0; }
I would use a loop like this: const char *p= str-1; size_t count= 0; while (*++p) if (islower (*p)) ++count;
int count_whitespace (FILE* input) { int c, count=0; while (( c = getc(input) ) != EOF ) if ((char) c==' ') ++count; return count; }
int i; for (i=1; i<=10; i++) printf ("%d %d\n", i, i*i);
in linux wc -l filename will count the lines and wc will count the letters
there could be a part in it like this: int num, digit; int count [10]; do { digit = num%10; num != 10; ++count[digit]; } while (num);
too many to count
To write repeated multiplication in an exponential notation, you should write the number that has to be multiplied as the base. Count the number of times that the number is used.