#include<stdio.h>
void main()
{
int n,sum=0;
clrscr();
printf("enter the value"):
scanf("%d",&n);
do
{
sum=sum+n;
n=n-1;
printf("sum is %d"<sum);
}
while(n>0)
getch();
}
Well, it's very hard to write a flowchart in text, so I'll give you some pseudo code instead. int number = the given number int sum = 0 loop while number is not 0 sum = sum + (number mod 10) number = number / 10
There are three forms of loop commonly used in C/C++, the for loop, the while loop and the do-while loop. The for loop is most commonly used whenever an action is going to be performed a set amount of times. For example, to sum every element in an array: for(i = 0; i < arraySize; i++) { sum = sum + array[i]; } The while loop and do-while loop are commonly used to loop until a condition is met. The difference between the two is that the do-while loop goes through one iteration before checking its condition, while the while loop checks its condition before any execution of the loop. Example do-while loop: do { randomNumber = rand() % 10; }while(randomNumber != 6); Example while loop: cout > number; while(number < 0) { cout > number; }
int i, sum; /* example using while */ sum = 0; i = 1; while (i <= 100) { sum += i; ++i; } /* example using for */ sum = 0; for (i = 1; i <= 100; ++i) sum += i;
import java.util.*;public class Digit{public static void main(String[] args){Scanner sc = new Scanner(System.in);System.out.print("Enter Number:");String r = sc.nextLine();int[] b = new int[r.length()];String [] a = new String[r.length()];int k = 0;System.out.println("\n");System.out.print("The Digits of the number are: ");for (int i = 0; i < r.length(); i++){a[i] = r.substring(i,i+1);System.out.print(a[i]);System.out.print(" ");k = k + Integer.parseInt(a[i]);}System.out.println("\n");System.out.print("The sum of the digits is: ");System.out.println(k);}}Output:C:\>java DigitEnter Number:123456789The Digits of the number are: 1 2 3 4 5 6 7 8 9The sum of the digits is: 45Modified while loopAnswer// start off with an integer int n = some_integer;// this will keep track of the sum of the digits of n...// we set it to n initially to simplify our loop logicint sum_digits = n;// this loop is to handle cases where the sum of the digits is initially greater than 10// and we need to sum the digits againwhile(sum_digits >=10) {// reset sum_digitssum_digits = 0;// use log function to get the number of digits in nint num_digits = (int) Math.log10(n) + 1;// here is the actual work of the algorithm...do {// (n % 10) will return the last (right-most) digit of n,// which we will add to our sumsum_digits += n % 10;// now divide n by 10 to get rid of the digit we just addedn /= 10;}while( n > 0 ); // loop until we're done with this number// set n to the sum of the digits in case we need to loop againn = sum_digits}
class sum { void main () { int sum = 0; int n = 1; while ( n <= 100 ) { sum = sum + n; n++ ; } System.out.println("Sum is = " + sum ); }}
In python 3 I would write the following (assuming the digits are input by a user): #!/usr/bin/python a = input("Type a number: ") sum = 0 for i in a: #the for next loop performs the function of a while loop with less lines of code. #(My formating keeps getting deleted when I publish this answer. There should always be a tab- strictly speaking four spaces before this line) sum = sum + int(i) sum = str(sum) print("Your number was", a + ". The sum of its digits is" , sum + ".") As you can see, the for next loop iterates through each value in the string "a", and sets its value to "i". For each iteration, the current value of i is added to the variable "sum" until the end of the number is reached. The program then prints the original number, and the sum of its digits.
#include using std::cout;using std::endl;int main(){int number(1);cout number;//using loop forint sum(0);for(int i(1); i
Well, it's very hard to write a flowchart in text, so I'll give you some pseudo code instead. int number = the given number int sum = 0 loop while number is not 0 sum = sum + (number mod 10) number = number / 10
Using for loop we can find sum of digits of a number. Inside the loop just perform Logic Expression 1) rem=num%10. {To find the unit place no. using remainder functon} 2) sum = sum+rem {to find the addition ie output} 3) num=num/10 {to eliminate the added digit} Just repeat these steps in the loop.
Place the digits in an array (or vector) and use the loop to sum each digit to a running total initialised to zero. The following function demonstrates the loop: int sum_vector(std::vector<int>& v) { int total=0; int i=0; while( i<v.size() ) total+=v[i]; return total; }
the sum of my digits is 6? answer=60
Add the digits together. The sum of the digits of 23 is 5.
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
class Sum_Of_Digits { public static void printSumandnoofdigits(int n) { int temp = n; int count = 0; int sum = 0; while ( n > 0 ) { sum = sum + n % 10; n = n / 10; count ++; } System.out.println("The number is..." + temp ); System.out.println("The sum of digits is..." + sum); System.out.println("The number of digits is..." + count); } }
There are three forms of loop commonly used in C/C++, the for loop, the while loop and the do-while loop. The for loop is most commonly used whenever an action is going to be performed a set amount of times. For example, to sum every element in an array: for(i = 0; i < arraySize; i++) { sum = sum + array[i]; } The while loop and do-while loop are commonly used to loop until a condition is met. The difference between the two is that the do-while loop goes through one iteration before checking its condition, while the while loop checks its condition before any execution of the loop. Example do-while loop: do { randomNumber = rand() % 10; }while(randomNumber != 6); Example while loop: cout > number; while(number < 0) { cout > number; }
8 is the same as the sum of the digits of its cube, 512.
1 + 1 = 2 The sum of the digits is therefore 2.