1.8 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f45b1000cf542c50ff6d | Problem 238: Infinite string tour | 5 | 301883 | problem-238-infinite-string-tour |
--description--
Create a sequence of numbers using the "Blum Blum Shub" pseudo-random number generator:
s_0 = 14025256 \\\\
s_{n + 1} = {s_n}^2 \\; mod \\; 20\\,300\\,713
Concatenate these numbers s_0s_1s_2\ldots
to create a string w
of infinite length. Then, w = 14025256741014958470038053646\ldots
For a positive integer k
, if no substring of w
exists with a sum of digits equal to k
, p(k)
is defined to be zero. If at least one substring of w
exists with a sum of digits equal to k
, we define p(k) = z
, where z
is the starting position of the earliest such substring.
For instance:
The substrings 1, 14, 1402, … with respective sums of digits equal to 1, 5, 7, … start at position 1, hence p(1) = p(5) = p(7) = \ldots = 1
.
The substrings 4, 402, 4025, … with respective sums of digits equal to 4, 6, 11, … start at position 2, hence p(4) = p(6) = p(11) = \ldots = 2
.
The substrings 02, 0252, … with respective sums of digits equal to 2, 9, … start at position 3, hence p(2) = p(9) = \ldots = 3
.
Note that substring 025 starting at position 3, has a sum of digits equal to 7, but there was an earlier substring (starting at position 1) with a sum of digits equal to 7, so p(7) = 1
, not 3.
We can verify that, for 0 < k ≤ {10}^3
, \sum p(k) = 4742
.
Find \sum p(k)
, for 0 < k ≤ 2 \times {10}^{15}
.
--hints--
infiniteStringTour()
should return 9922545104535660
.
assert.strictEqual(infiniteStringTour(), 9922545104535660);
--seed--
--seed-contents--
function infiniteStringTour() {
return true;
}
infiniteStringTour();
--solutions--
// solution required