(HackerRank) Write a Modular C Programming code to solve Divisor summation 1, Given a natural number n output the summation of all its proper divisors.
Given a natural number n output the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number. e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
Input Format
Natural number n
Constraints
n should be positive
Output Format
Divisor summation of n
Sample Input 0
20
Sample Output 0
Divisor summation is 22
Sample Input 1
-5
Sample Output 1
Invalid natural number
Refer : C Programming HackerRank all solutions for Loops | Arrays | strings
CODE:
#include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> void divisor(int n) { int sum=0,i; for(i=1;i<n;i++) { if(n%i==0) { sum=sum+i; } } printf("Divisor summation is %d\n",sum); } int main() { int n; scanf("%d",&n); if(n<1) { printf("Invalid natural number\n"); exit(0); } else { divisor(n); } return 0; }
OUTPUT
Congratulations, you passed the sample test case. Click the Submit Code button to run your code against all the test cases. Input (stdin) 20 Your Output (stdout) Divisor summation is 22 Expected Output Divisor summation is 22
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