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C Program to check Plus Perfect Number
Given a number x with n number of digits, our task is to check whether the given number’s Plus Perfect number or not. In order to check that the number is Plus Perfect Number we find the nth power of every digit d (d^n) and then sum all the digits, if the sum is equal to n then the number is Plus Perfect Number. Plus Perfect number is similar like finding an Armstrong of any number.
Like In the given example below −
Example
Input: 163 Output: Number is not a perfect_number Explanation: 1^3 + 6^3 + 3^3 is not equal to 163 Input: 371 Output: Number is a perfect_number Explanation: 3^3 + 7^3 +1^3 is equal to 371
The approach used below is as follows −
- First step is to count the number of digits in the input given.
- Second step is to power a digit the same number of times as the number of digits of the input.
- Third step is adding all the numbers and check if it is equal or not.
Algorithm
Start In function int power(int a, int b) Step 1-> Declare and initialize power as 1 Step 2-> Loop While b>0 Set power = power * a Decrement b by 1 Step 3-> return power End function power In function int count(int n) Step 1-> Declare and Initialize i as 0 Step 2-> Loop While n!=0 Increment i by 1 Set n = n/10 End Loop Step 3-> Return i In function int perfect_number(int n) Step 1-> Declare and initialize x as count(n) Step 2-> Declare and initialize rem as 0 and m as 0 Step 3-> Loop While(n) Set rem as n %10 Set m as m + power(rem, x) Set n as n/ 10 End Loop Step 4-> Return m End Function perfect_number In Function int main(int argc, char const *argv[]) Step 1-> Initialize n as 1634 Step 2-> If n == perfect_number(n) then, Print "Number is a perfect_number " Step 3-> else Print "Number is not a perfect_number " End if End main Stop
Example
#include <stdio.h>
int power(int a, int b) {
int power =1;
while(b>0) {
power *= a;
b--;
}
return power;
}
int count(int n) {
int i=0;
while(n!=0) {
i++;
n = n/10;
}
return i;
}
int perfect_number(int n) {
int x = count(n);
int rem = 0, m=0;
while(n) {
rem = n %10;
m += power(rem, x);
n /= 10;
}
return m;
}
int main(int argc, char const *argv[]) {
int n = 1634;
if(n == perfect_number(n)) {
printf("Number is a perfect_number
");
}
else
printf("Number is not a perfect_number
");
return 0;
}Output
If run the above code it will generate the following output −
Number is a perfect_number
Updated on: 21-Oct-2019
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