1.3 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3df1000cf542c50fef1 | Problem 115: Counting block combinations II | 5 | 301741 | problem-115-counting-block-combinations-ii |
--description--
A row measuring n
units in length has red blocks with a minimum length of m
units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.
Let the fill-count function, F(m, n)
, represent the number of ways that a row can be filled.
For example, F(3, 29) = 673135
and F(3, 30) = 1089155
.
That is, for m = 3, it can be seen that n = 30 is the smallest value for which the fill-count function first exceeds one million.
In the same way, for m = 10, it can be verified that F(10, 56) = 880711
and F(10, 57) = 1148904
, so n = 57 is the least value for which the fill-count function first exceeds one million.
For m = 50, find the least value of n
for which the fill-count function first exceeds one million.
Note: This is a more difficult version of Problem 114.
--hints--
countingBlockTwo()
should return 168
.
assert.strictEqual(countingBlockTwo(), 168);
--seed--
--seed-contents--
function countingBlockTwo() {
return true;
}
countingBlockTwo();
--solutions--
// solution required