int sum_digits (int num) {
int sum;
sum =0;
while (num) {
sum += num % 10;
num /= 10;
}
return sum;
}
Example usage:
int main (void) {
int i;
printf ("Input a 6-digit number: ");
scanf ("%6d", &i);
printf ("The sum of the digits in %d is: %d\n", i, sum_digits (i));
return 0;
}
size_t sum_of_digits (size_t num)
{
size_t sum {0};
for ( ; num != 0; num /= 10)
{
sum += num % 10;
}
return sum;
}
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Reference:cprogramming-bd.com/c_page2.aspx# extract the digit
Function sum_odd_digits(ByVal number As Integer) As Integer Dim digit As Integer sum_odd_digits = 0 While number <> 0 digit = number Mod 10 If digit And 1 Then sum_odd_digits = sum_odd_digits + digit number = number / 10 End While End Function
Add the last digit plus the sum of all the previous digits. The base case is that if your integer only has a single digit, just return the value of this digit. You can extract the last digit by taking the remainder of a division by 10 (number % 10), and the remaining digits by doing an integer division by 10.
The tricky part is getting the individual digits. There are basically two ways to do this: 1) Convert the number to a string, and use string manipulation to get the individual digits. 2) Repeatedly divide the number by 10. The digit is the remainder (use the "%" operator). To actually get the highest digit, initially assume that the highest digit is zero (store this to a variable, called "maxDigit" or something similar). If you find a higher digit, replace maxDigit by that.
Use an enum if you are using a c style language. Or a map data structure. Assign each integer an English value and then match it to what the user inputs.
In a three digit integer number the tenthsdigit is always 0 as integer numbers are whole numbers and have no decimal part and tenths are decimal parts:tenths_digit_of_integer_number = 0I suspect you mean "How to find the tens digit of an integer number?"; this is the second from the right, so:tens_digit = (INT(number ÷ 10)) MOD 10For example, in C this would become: tens_digit = (number / 10) % 10;
It is -987. The smallest positive 3-digit integer with unique digits is 102.
Input the number as a string. If the string has a length of 4 and contains only digits, convert the string to an integer. If the integer is less than 1000, input another number. Otherwise, copy the integer and divide by 100 to get rid of the two least-significant digits. Divide again by 2 and take the remainder. If the remainder is 1 then the second left digit of the 4-digit integer is odd and the 4-digit integer can be added to the array, otherwise do not add it. Repeat for all n numbers.
Function sum_odd_digits(ByVal number As Integer) As Integer Dim digit As Integer sum_odd_digits = 0 While number <> 0 digit = number Mod 10 If digit And 1 Then sum_odd_digits = sum_odd_digits + digit number = number / 10 End While End Function
To count the number of times a digit occurs in an integer, start by initializing an array of ten counts of digits, such as int digits[10];Then, in a loop while the number is non zero, increment the element in the digits array that corresponds to the units digit, and then divide the number by ten, such as digits[number%10]++ and number/=10;int digits[10];int i;int number = some number;for (i=0; i
All digits all part of the set of integers.
15
The answer is 9*9!/9*109 = 0.0003629 approx.
Normally a 2-digit number refers to an integer with two digits, the first of which is not 0. So the answer would be NO> But it is a number with 2 significant digits.
81
A number with more than one digits: that is, an integer greater than 9.
Because the next larger integer is 1000000 which has 7 digits.
The answer is a negative or positive integer with one or two digits.