Dark Mode On/Off

Oddly Even:

Given a maximum of 100 digit numbers as input, find the difference between the sum of odd and even position digits.

Case 1:

Input: 4567

Expected Output: 2

Explantion : Odd positions are 4 and 6 as they are pos: 1 and pos: 3, both have sum 10. Similarly, 5 and 7 are at even positions pos: 2 and pos: 4 with sum 12. Thus, difference is 12 - 10 = 2

Given a maximum of 100 digit numbers as input, find the difference between the sum of odd and even position digits.

Case 1:

Input: 4567

Expected Output: 2

Explantion : Odd positions are 4 and 6 as they are pos: 1 and pos: 3, both have sum 10. Similarly, 5 and 7 are at even positions pos: 2 and pos: 4 with sum 12. Thus, difference is 12 - 10 = 2

// Find the Nth term in the series

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

int main()

{

int a = 0,b = 0,i = 0, n;

char num[100];

printf("Enter the number:");

scanf("%s",num); //get the input up to 100 digit

n = strlen(num);

while(n>0)

{

if(i==0) //add even digits when no of digit is even and vise versa

{

a+=num[n-1]-48;

n--;

i=1;

}

else //add odd digits when no of digit is even and vice versa

{

b+=num[n-1]-48;

n--;

i=0;

}

}

printf("Difference between the sum of odd and even position digits is %d",abs(a-b)); //print the difference of odd and even

return 0;

}

© 2024 Studytonight Technologies Pvt. Ltd.