Question

An array contains 2 or 3 digit positive numbers (9<n<1000).Write a program which counts the number of numbers in this array having their leftmost digit equal to their rightmost digit (examples: 44,66,202,767,777....)

Write a code by using c programming.Not c++

Answer 1

Source code in C:

#include <stdio.h> int main() { /*required variables*/ int size, leftmost=0, rightmost=0,count=0,i; /*read size from the user*/ printf("Enter the size of the array: "); scanf("%d", &size); /*array declaration*/ int array[size]; /*read numbers to the array*/ printf("Enter the 2 or 3 digit numbers:\n"); for(i=0; i<size; i++) { /*read number from user*/ scanf("%d",&array[i]); /*check for 2 or 3 digit number*/ if(array[i]<10||array[i]>999) { /*if user entered a invalid number read again*/ i--; printf("Invalid number!!!enter again.. \n"); }

/*if user enters a valid number*/ else { /*if number is 2 digit number*/ if(array[i]<100) { /*find rightmost and leftmost*/ rightmost=array[i]%10; leftmost=array[i]/10; } /*if number is 3 digit number*/ else { /*find rightmost and Leftmost*/ rightmost=array[i]%10; leftmost=array[i]/100; } /*if both are equal increment count*/ if(rightmost==leftmost) { count++; } } }

/*display count*/ printf("The total number of numbers in the array"); printf("\nhaving leftmost digit equal to rightmost digit are: %d", count); return 0; }

Output:

C:\Users\Kumar Reddi\Desktop\LERD.exe Enter the size of the array: 10 Enter the 2 or 3 digit numbers: 27 44 58 66 98 202 574 767 884 777 The total number of numbers in the array having leftmost digit equal to rightmost digit are: 5 Process exited after 34.04 seconds with return value o Press any key to continue

Source code in code snippet format(text):

 #include <stdio.h> int main() {         /*required variables*/  int size,leftmost=0,rightmost=0,count=0,i;      /*read size from the user*/     printf("Enter the size of the array: ");        scanf("%d",&size);  /*array declaration*/   int array[size];        /*read numbers to the array*/   printf("Enter the 2 or 3 digit numbers:\n");    for(i=0;i<size;i++)  {               /*read number from user*/               scanf("%d",&array[i]);              /*check for 2 or 3 digit number*/               if(array[i]<10||array[i]>999)             {                       /*if user entered a invalid number read again*/                         i--;                    printf("Invalid number!!!enter again..\n");             }               /*if user enters a valid number*/               else            {                       /*if number is 2 digit number*/                         if(array[i]<100)                     {                               /*find rightmost and leftmost*/                                 rightmost=array[i]%10;                          leftmost=array[i]/10;                   }                       /*if number is 3 digit number*/                         else                    {                               /*find rightmost and leftmost*/                                 rightmost=array[i]%10;                          leftmost=array[i]/100;                  }                       /*if both are equal increment count*/                   if(rightmost==leftmost)                         {                               count++;                        }               }       }       /*display count*/       printf("The total number of numbers in the array");     printf("\nhaving leftmost digit equal to rightmost digit are: %d",count);               return 0; }