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1. Compute sum of digits in all numbers from 1 to n
Input: 11
ouput: 66 (1+2+3+4+5+6+7+8+9+10+11)
Pattern
sum(9) = 1 + 2 + 3 + 4 ........... + 9
= 9*10/2
= 45
sum(99) = 45 + (10 + 45) + (20 + 45) + ..... (90 + 45)
= 45*10 + (10 + 20 + 30 ... 90)
= 45*10 + 10(1 + 2 + ... 9)
= 45*10 + 45*10
= sum(9)*10 + 45*10
sum(999) = sum(99)*10 + 45*100
sum(10d - 1) = sum(10d-1 - 1) * 10 + 45*(10d-1)
Algorithm
1) Find number of digits minus one in n. Let this value be 'd'.
For 328, d is 2.
2) Compute some of digits in numbers from 1 to 10d - 1.
Let this sum be w. For 328, we compute sum of digits from 1 to
99 using above formula.
3) Find Most significant digit (msd) in n. For 328, msd is 3.
4) Overall sum is sum of following terms
a) Sum of digits in 1 to "msd * 10d - 1". For 328, sum of
digits in numbers from 1 to 299.
For 328, we compute 3*sum(99) + (1 + 2)*100. Note that sum of
sum(299) is sum(99) + sum of digits from 100 to 199 + sum of digits
from 200 to 299.
Sum of 100 to 199 is sum(99) + 1*100 and sum of 299 is sum(99) + 2*100.
In general, this sum can be computed as w*msd + (msd*(msd-1)/2)*10d
b) Sum of digits in msd * 10d to n. For 328, sum of digits in
300 to 328.
For 328, this sum is computed as 3*29 + recursive call "sum(28)"
In general, this sum can be computed as msd * (n % (msd*10d) + 1)
+ sum(n % (10d))
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