For example take 2345, how we add middle numbers 3 and 4 and gives output ..the sum of middle numbers is 7.
xavier This code works for 4 digit number. If you want code that works for any digit, you can tell me. I will give you. #include int main[] { int a,sum; printf["Enter a number: "]; scanf["%d",&a]; sum=[a/10]%10+[a/100]%10; printf["Answer: %d",sum]; return 0; }
The easiest way is to convert the number into an array of characters, take the substring and then find the sum.
i have tried but no use can you provide me with complete code
If you are sure it is a four digit number then #include /* Function to get sum of digits */ int getSum[int n] { int sum; sum=0; /* Single line that calculates sum */ sum=sum+[n/100]%10+[n%100]/10; return sum; } // Driver code int main[] { int n = 3547; printf[" %d ", getSum[n]]; return 0; }
Can you suugest me with code
/* Function to get sum of digits */ int getSum[int n] { int sum; /* Single line that calculates sum */ for [sum = 0; n > 0; sum += n % 10, n /= 10] ; return sum; } // Driver code int main[] { int n = 3547; printf[" %d ", getSum[n]]; return 0; }
Use while loop to get number of digits
i dont know to use while loop that is why i request you to provide code
Am new here who can teach me python directly
Given a number, find the sum of its digits.
Examples :
Input: n = 687
Output: 21Input: n = 12
Output: 3
Follow the below steps to solve the problem:
- Get the number
- Declare a variable to store the sum and set it to 0
- Repeat the next two steps till the number is not 0
- Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
- Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
- Print or return the sum
Below is the implementation of the above approach:
C++
#include
using namespace std;
class gfg {
public:
int getSum[int n]
{
int
sum = 0;
while [n != 0] {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
};
int main[]
{
gfg g;
int n = 687;
cout
Javascript
function getSum[n]
{
var sum = 0;
while [n != 0] {
sum = sum + n % 10;
n = parseInt[n / 10];
}
return sum;
}
var n = 687;
document.write[getSum[n]];
Time Complexity: O[log N]
Auxiliary Space: O[1]
How to compute in a single line?
The below function has three lines instead of one line, but it calculates the sum in one line using for loop. It can be made one-line function if we pass the pointer to the sum.
Below is the implementation of the above approach:
C++
#include
using namespace std;
class gfg {
public:
int getSum[int n]
{
int
sum;
for [sum = 0; n > 0; sum += n % 10, n /= 10]
;
return sum;
}
};
int main[]
{
gfg g;
int n = 687;
cout 0; sum += n % 10, n /= 10]
;
return sum;
}
int main[]
{
int n = 687;
printf[" %d ", getSum[n]];
return 0;
}
Java
import java.io.*;
class GFG {
static int getSum[int n]
{
int sum;
for
[sum = 0; n > 0; sum += n % 10, n /= 10]
;
return sum;
}
public static void main[String[] args]
{
int n = 687;
System.out.println[getSum[n]];
}
}
Python3
def getSum[n]:
sum
= 0
while[n > 0]:
sum += int[n % 10]
n = int[n/10]
return sum
if __name__ == "__main__":
n = 687
print[getSum[n]]
C#
using System;
class GFG {
static
int getSum[int n]
{
int sum;
for [sum = 0; n > 0; sum += n % 10, n /= 10]
;
return sum;
}
public static void Main[]
{
int n = 687;
Console.Write[getSum[n]];
}
}
PHP
Javascript
function getSum[n]
{
let sum;
for[sum = 0; n > 0;
sum += n % 10,
n = parseInt[n / 10]]
;
return
sum;
}
let n = 687;
document.write[getSum[n]];
Time Complexity: O[log N]
Auxiliary Space: O[1]
Sum of the digits of a given number using recursion:
Follow the below steps to solve the problem:
- Get the number
- Get the remainder and pass the next remaining digits
- Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
- Divide the number by 10 with help of the ‘/’ operator to remove the rightmost digit.
- Check the base case with n = 0
- Print or return the sum
Below is the implementation of the above approach:
C++
#include
using namespace std;
class gfg {
public:
int sumDigits[int no]
{
if
[no == 0] {
return 0;
}
return [no % 10] + sumDigits[no / 10];
}
};
int main[void]
{
gfg g;
cout
Javascript
function sumDigits[no]
{
if[no == 0]{
return 0 ;
}
return [no % 10] + sumDigits[parseInt[no/10]] ;
}
document.write[sumDigits[687]];
Time Complexity: O[log N]
Auxiliary Space: O[log N]
Sum of the digits of a given number with input as string:
When the number of digits of that number exceeds 1019 , we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character[in this case it is numeric]
Follow the below steps to solve the problem:
- Declare a variable sum equal to zero
- Run a loop from zero to the length of the input string
- Add the value of each character into the sum, by converting the character into it’s integer value
- Return sum
Below is the implementation of the above approach:
C++14
#include
using namespace std;
int getSum[string str]
{
int sum = 0;
for [int i = 0; i < str.length[]; i++] {
sum = sum + str[i] - 48;
}
return sum;
}
int main[]
{
string st = "123456789123456789123422";
cout
Javascript
function getSum[str]
{
let sum = 0;
for [let i = 0; i < str.length; i++]
{
sum = sum + parseInt[str[i]];
}
return sum;
}
let st = "123456789123456789123422";
document.write[getSum[st]];
Time Complexity: O[log N]
Auxiliary Space: O[1]
Sum of the digits of a given number using tail recursion:
Follow the below steps to solve the problem:
- Add another variable “Val” to the function and initialize it to [ Val = 0 ]
- On every call to the function add the mod value [n%10] to the variable as “[n%10]+val” which is the last digit in n. Along with passing the variable n as n/10.
- So on the First call, it will have the last digit. As we are passing n/10 as n, It follows until n is reduced to a single digit.
- n