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Given two integers "number" and "repeat" as input. The goal is to calculate the numerical sum of an input number repeated a "repeat" number of times until the sum becomes a single number. Continue in this way until the sum of the numbers obtained becomes a single digit. If the input number is 123 and repeat=2 then the sum of 123123 will be 1 2 3 1 2 3=12 This is not a single digit. Now the sum of the digits of 12 is 1 2 = 3. The output will be 3
Input− number=32 Repeat=3
Output− The recursive sum of the numbers formed by repeated appending is: 6
Explanation − The sum of the numbers of 323232 is 3 2 3 2 3 2=15, and the sum of the digits of 15 is 1 5=6. 6 is a single digit, so the output will be 6.
Input− number=81 Repeat=4
Output−The recursive digital sum of the numbers formed by repeated appending is: 9
Explanation - The sum of the digits of 81818181 is 1 8 1 8 1 8 1 8=36, and the sum of the digits of 36 is 3 6=9. 9 is a single digit, so the output will be 9.
Declare two integer type variables such as number and repeat. Pass the data to the function as Recursive_Sum(number, Repeat).
Inside the function as Recursive_Sum(int number, int Repeat)
Declare an integer variable as total and use repeat * sum(number ) sets it;
Returns a call to the function as sum(total) .
Inside the function as sum (int number)
Check if the IF number is 0 and return 0.
Checks if the IF number % 9 is 0 and returns 9.
ELSE, return number % 9
Print the result.
#include <bits/stdc++.h> using namespace std; int sum(int number){ if(number == 0){ return 0; } if(number % 9 == 0){ return 9; } else{ return number % 9; } } int Recursive_Sum(int number, int repeat){ int total = repeat * sum(number); return sum(total); } int main(){ int number = 12; int repeat = 4; cout<<"Recursive sum of digits of a number formed by repeated appends is: "<<Recursive_Sum(number, repeat); return 0; }
If we run the above code it will generate the following output
Recursive sum of digits of a number formed by repeated appends is: 3
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