C programming

Write a C program to output the following multiplication table of a number,when the inputs are keyed in from the keyboard.

5x1=5
5x2=10
5x3=15
-------
-------
5x10=50

 

Code:

#include<conio.h>
#include<stdio.h>
main()
{
    float a,b,c;
    clrscr();
    printf("enter a and b values\n");
    scanf("%f%f",&a,&b);
    c=a*b;
    printf("Multiplication of a and b is:%f",c);
    getch();
}

 

Output:

enter a and b values
5
1
Multiplication of a and b is: 5

//i.e., the number and multiplier , we asume it as a and b

 

#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a,b;
printf("enter value");
scanf("%d", &a);
printf("enter value again");
scanf("%d", &b);
printf("%d + %d = %d\n", a, b, a*b);
getch();
}


output
enter value
5
enter another value
1
5*1=5

 

For multiplication table of a number,when the inputs are keyed in from the keyboard:

#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a,b;
printf("enter the values");
scanf("%d", &a);
scanf("%2d", &b);
while(b<=10)
{
printf("%d * %2d = %2d\n", a, b, a*b);
b=b+1;
}
getch();
}

output
enter the values
5
1
5*1=5
5*2=10
5*3=15
5*4=20
5*5=25
...........
...........
5*10=50

 

Code:

#include <stdio.h>
#include <string.h>
#include <limits.h>

int main(void) {

   char buf[10] = { 0 };
   int i, num;

   do {

      /* wait for valid numerical input */

      printf("Enter table multiplier: ");
      fgets(buf, sizeof buf, stdin);

   } while((sscanf(buf, "%d", &num)) != 1);

   for(i=1; (i * num) < INT_MAX; ++i) {

      if(i % 10 == 0) {
	     puts("\npress enter");
	     getchar(); // pause output
      }

      printf("%3d %2c %3d = %3d\n", num, 'X', i, i * num);
   }

   return 0;
}

:P