for 循环和模运算符

Question

Write a program that calculates the sum of the digits of an integer. For example, the sum of the digits of the number 2155 is 2 + 1 + 5 + 5 or 13. The program should accept any arbitrary integer typed in by user.

I can get it working using while loop but if I sub it w/ a for loop, the program calculates all of numbers except the last. Why is this?

#include <stdio.h>

int main(void)
{
    int i, num, sum = 0, temp;

    printf("Enter the number:\n");
    scanf("%i", &num);

    printf("Test for num: %i\n", num); //both num & temp return same number
    temp = num;
    printf("Test for temp: %i\n", temp);

    //while (num > 0)
    for (i = 0; i <= temp; i++)  //replacing temp w/ num will skip last num
    {
        sum += num % 10;
        //sum += right;
        num /= 10;
    }

    printf("Given number = %i\n", temp);
    printf("Sum of digits of %i = %i", temp, sum);

    return (0);
}

Answer 1

With num in the for loop as you have commented out, you are counting i against the dividend of the original number, not while num > 0 .

If you have, for example, num = 158, the loop would execute, then set num to 15. i is incremented to 1. Thus i < num, so it executes again. After the loop this time, num == 1 and i == 2. Hence it will not execute and the 1 of 158 is not added.

If your highest digit is greater than or equal to the number of digits, your code with num in the for loop would work. Otherwise, it will not.

You can get rid of i and simply use num in your for loop.

for(;num > 0; num /= 10)
    sum += num%10;

Answer 2

Note:

for (i = 0; i <= temp; i++)

It is not fair - if temp is for example 543 , you certainly don't perform this loop 544 times (in spite the result is OK as the loop in its majority iterations only adds 0 to already correct result).

Your program with its original while loop

while (num > 0)
{
    sum += num % 10;
    num /= 10;
}

works OK for relatively small numbers , ie in the int range ^*) , I tested it, for example

Enter the number: 1234 Test for num: 1234 Test for temp: 1234 Given number = 1234 Sum of digits of 1234 = 10

or

Enter the number: 123456789 Test for num: 123456789 Test for temp: 123456789 Given number = 123456789 Sum of digits of 123456789 = 45

But, for example

Enter the number: 10000000001 Test for num: 1410065409 Test for temp: 1410065409 Given number = 1410065409 Sum of digits of 1410065409 = 30

You may see that the scanf() function read the "big" number 10000000001 as 1410065409 !

But it is not a problem with the logic of your while loop, the result for the number 1410065409 is correct.

---

(*) - int range for the most common implementations of int (as 32 bits numbers) is

              from  -2.147.483.648  to  +2.147.483.647.

Answer 3

Do this, print out the variable i in side your for loop and see how often it runs. This is inefficient and a clear waste of resources.

You should also be considering the following ?

What is the time complexity? What is different about that of the while loop vs your for loop using temp?

When you changed to a for loop you did not consider what was happening the variable num in the while loop. Consider, a number with n digits is between 10^(n-1) inclusive and 10^n exclusive. If we let n be the number of digits of in N, then the inequality is 10^(n-1) <= N < 10^n . From this we find that the time complexity is O(log(n)). There are roughly log10(num) digits in num.

Your solution is correct in that is produces the correct answer but is ineffecient in performance. Firstly, you should be reducing the for loop index as such.

 for (i = temp ; i !=0; i /= 10)

This would be more correct using a for loop. This would run the same number of times as the while loop, but would need to decrement i and check if i != 0 for iteration.