684 Inverse Digit Sum.sf

#!/usr/bin/ruby

# Daniel "Trizen" Șuteu
# Date: 27 November 2019
# https://github.com/trizen

# See OEIS sequence:
#   https://oeis.org/A051885

# The smallest numbers whose sum of digits is n, are numbers of the form r*10^j-1, with r=1..9 and j >= 0.

# This solution uses the following formula:
#   Sum_{j=0..n} (r*10^j-1) = (r * 10^(n+1) - r - 9*n - 9)/9

# By letting r=1..9, we get:
#   R(k) = Sum_{r=1..9} Sum_{j=0..n} (r*10^j-1) = 2*(2^n * 5^(n+2) - 7) - 9*n

# From R(k), we get S(k) as:
#   S(k) = R(k) - Sum_{j=2+(k mod 9) .. 9} (j*10^n-1)

# where:
#   n = floor(k/9)

# https://projecteuler.net/problem=684

# Runtime: 0.188s

const MOD = 1000000007

func S(k) {

    var n   = floor(k/9)
    var sum = (2*(2**n * 5**(n+2) - 7) - 9*n)

    for r in (k%9 + 2 .. 9) {
        sum -= (r * 10**n - 1)
    }

    sum
}

func modular_S(k) {

    var n   = floor(k/9)
    var sum = (2*(powmod(2, n, MOD) * powmod(5, n+2, MOD) - 7) - 9*n)%MOD

    for r in (k%9 + 2 .. 9) {
        sum -= (r * powmod(10, n, MOD) - 1)%MOD
    }

    sum % MOD
}

assert_eq(S(20), 1074)
assert_eq(S(49), 1999945)

assert_eq(modular_S(20), 1074)
assert_eq(modular_S(49), 1999945)

say sum(2..90, {|k|
    modular_S(fib(k))
})%MOD