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Given two integers A and B , how to efficiently compute the sum of all the digits of every number int the set $\{N | A \le N \le B \} $.

For example: A = 100 and B = 777, then the required answer is 8655.

I am interested in deriving an formula/efficient algorithm for the same.

3 Answers 3

7

You should find the formula $f(n)$ for the sum of digits for numbers from 1 to n, then do $f(B)-f(A)$. It is not very clean. The sum of digits of all k digit numbers, that is from $10^{(k-1)} \text { to } 10^k-1$, is $45*(10^k-1)/9$. Stopping part way is harder, but you can do it by recursion. If you want the sum of digits of numbers up to $7655$, say, you can do sum of digits up to $999$, plus $1000*6*5/2$ (for the thousands digits up through 6) + $6*$sum of digits up to $999$ (for the last three digits for $1000$ through $6999$) + $7*656$ (for the thousands in the $7$s) + sum of digits up to $655.$ The last has one less digit, so you just call your subroutine with that.

  • 0
    +1, seems a feasible approach but I haven't understood it fully,could you explain in lucid manner ?2010-11-02
  • 0
    If you look at the answer to http://math.stackexchange.com/questions/8576/number-of-digits-from-1-to-n/8577#8577 you will see a more detailed explanation for the total number of digits in the n digit numbers. Then you have to deal with the partial decade. Do you have a specific question on that?2010-11-02
  • 0
    There's more information in Sloane's A037123.2011-06-03
  • 0
    @Quixotic see this for a part of your solution. http://www.geeksforgeeks.org/count-sum-of-digits-in-numbers-from-1-to-n/2016-09-09
2

Code in C++

#include
#include
#include
using namespace std;
long long  b,i,x,v,c,ans,anc;
int j,k,a[20];
main(){

    // freopen("a.in","r",stdin);
   //  freopen("a.out","w",stdout);

     cin>>b;
     k=0; x=1; j=0;
     while(b!=0) {a[k]=b%10; b/=10; k++;} for (i=1; i>b;
      k=0; x=1; j=0;
     while(b!=0) {a[k]=b%10; b/=10; k++;} for (i=1; i
  • 1
    Your code seems to work fine on my machine, but perhaps you could add an explanation as to why it works?2012-12-22
  • 0
    Also FYI, this question appears to be a Spoj contest problem (see [this](http://www.spoj.com/problems/CPCRC1C/)).2012-12-22
-1

Just summing them altogether first would produce a number that has the same digit-sum as stringing them altogether.

For example: 101-103

101, 102, 103 -> 1+0+1+1+0+2+1+0+3 -> 9

101+102+103 = 306 = 3+0+6 = 9

Use the Gauss method to add the numbers quickly... That is, if I have the numbers in a set like 1-10 1+2+3+4+5+6+7+8+9+10 = 1+10+2+9+3+8+4+7+5+6 = (1+10)+(2+9)+(3+8)+(4+7)+(5+6) = (11)+(11)+(11)+(11)+(11) = 11*(10/2) <--- 10 is the upper-limit of the given range.

Then add the digits together by dividing by 10 and adding the remainder to one storing variable. (ex. 52 = 5+2 = 7.... 52 --> 52/10 --> 5r2... "2"+? .... 5/10 = 0r5 --> "2"+"5" = 7.)