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fix(curriculum): clean-up Project Euler 321-340 (#42988)

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* fix: clean-up Project Euler 321-340

* fix: typo

* fix: corrections from review

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>

* fix: corrections from review

Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>
Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
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2021-07-29 20:59:06 +02:00
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commit 1af6e7aa5a
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curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-325-stone-game-ii.md
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@ -8,9 +8,9 @@ dashedName: problem-325-stone-game-ii
# --description--
A game is played with two piles of stones and two players. At her turn, a player removes a number of stones from the larger pile. The number of stones she removes must be a positive multiple of the number of stones in the smaller pile.
A game is played with two piles of stones and two players. On each player's turn, the player may remove a number of stones from the larger pile. The number of stones removes must be a positive multiple of the number of stones in the smaller pile.
E.g., let the ordered pair(6,14) describe a configuration with 6 stones in the smaller pile and 14 stones in the larger pile, then the first player can remove 6 or 12 stones from the larger pile.
E.g., let the ordered pair (6,14) describe a configuration with 6 stones in the smaller pile and 14 stones in the larger pile, then the first player can remove 6 or 12 stones from the larger pile.
The player taking all the stones from a pile wins the game.
@ -18,16 +18,16 @@ A winning configuration is one where the first player can force a win. For examp
A losing configuration is one where the second player can force a win, no matter what the first player does. For example, (2,3) and (3,4) are losing configurations: any legal move leaves a winning configuration for the second player.
Define S(N) as the sum of (xi+yi) for all losing configurations (xi,yi), 0 &lt; xi &lt; yi ≤ N. We can verify that S(10) = 211 and S(104) = 230312207313.
Define $S(N)$ as the sum of ($x_i + y_i$) for all losing configurations ($x_i$, $y_i$), $0 &lt; x_i &lt; y_i ≤ N$. We can verify that $S(10) = 211$ and $S({10}^4) = 230\\,312\\,207\\,313$.
Find S(1016) mod 710.
Find $S({10}^{16})\bmod 7^{10}$.
# --hints--
`euler325()` should return 54672965.
`stoneGameTwo()` should return `54672965`.
```js
assert.strictEqual(euler325(), 54672965);
assert.strictEqual(stoneGameTwo(), 54672965);
```
# --seed--
@ -35,12 +35,12 @@ assert.strictEqual(euler325(), 54672965);
## --seed-contents--
```js
function euler325() {
function stoneGameTwo() {
return true;
}
euler325();
stoneGameTwo();
```
# --solutions--