* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
4.3 KiB
4.3 KiB
id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
5951815dd895584b06884620 | 给定半径的圆圈通过两个点 | 5 | circles-of-given-radius-through-two-points |
--description--
给定平面上的两个点和半径,通常可以通过这些点绘制给定半径的两个圆。
例外:零半径应视为从不描述圆(除非点是重合的)。如果这些点是重合的,则可以绘制无限数量的圆,其圆周上的点可以被绘制,除非半径也等于零,然后将圆圈折叠到一个点。如果点形成直径,则返回单个圆。如果这些点相距太远则无法绘制圆圈。任务:实现一个取两个点和一个半径的函数,并通过这些点返回两个圆。对于每个结果圆,提供每个圆的中心的坐标,四舍五入到四个十进制数字。将每个坐标作为数组返回,并作为数组数组进行坐标。对于边缘情况,请返回以下内容:如果点在直径上,则返回一个点。如果半径也为零,则返回"Radius Zero"
。如果点重合,则返回"Coincident point. Infinite solutions"
。如果点与直径相距更远,则返回"No intersection. Points further apart than circle diameter"
更远的"No intersection. Points further apart than circle diameter"
。样本输入: p1 p2 r 0.1234,0.9876 0.8765,0.2345 2.0 0.0000,2.000000 0.0000,0.0000 1.0 0.1234,0.9876 0.1234,0.9876 2.0 0.1234,0.9876 0.8765,0.2345 0.5 0.1234,0.9876 0.1234,0.9876 0.0参考:从数学论坛@Drexel的2点和半径中找到一个圆心
--hints--
getCircles
是一个函数。
assert(typeof getCircles === 'function');
getCircles([0.1234, 0.9876], [0.8765, 0.2345], 2.0)
应该返回[[1.8631, 1.9742], [-0.8632, -0.7521]]
。
assert.deepEqual(getCircles(...testCases[0]), answers[0]);
getCircles([0.0000, 2.0000], [0.0000, 0.0000], 1.0)
应该返回[0, 1]
assert.deepEqual(getCircles(...testCases[1]), answers[1]);
getCircles([0.1234, 0.9876], [0.1234, 0.9876], 2.0)
应返回Coincident point. Infinite solutions
assert.deepEqual(getCircles(...testCases[2]), answers[2]);
getCircles([0.1234, 0.9876], [0.8765, 0.2345], 0.5)
应返回No intersection. Points further apart than circle diameter
assert.deepEqual(getCircles(...testCases[3]), answers[3]);
getCircles([0.1234, 0.9876], [0.1234, 0.9876], 0.0)
应返回Radius Zero
assert.deepEqual(getCircles(...testCases[4]), answers[4]);
--seed--
--after-user-code--
const testCases = [
[[0.1234, 0.9876], [0.8765, 0.2345], 2.0],
[[0.0000, 2.0000], [0.0000, 0.0000], 1.0],
[[0.1234, 0.9876], [0.1234, 0.9876], 2.0],
[[0.1234, 0.9876], [0.8765, 0.2345], 0.5],
[[0.1234, 0.9876], [0.1234, 0.9876], 0.0]
];
const answers = [
[[1.8631, 1.9742], [-0.8632, -0.7521]],
[0, 1],
'Coincident point. Infinite solutions',
'No intersection. Points further apart than circle diameter',
'Radius Zero'
];
--seed-contents--
function getCircles(...args) {
return true;
}
--solutions--
const hDist = (p1, p2) => Math.hypot(...p1.map((e, i) => e - p2[i])) / 2;
const pAng = (p1, p2) => Math.atan(p1.map((e, i) => e - p2[i]).reduce((p, c) => c / p, 1));
const solveF = (p, r) => t => [parseFloat((r * Math.cos(t) + p[0]).toFixed(4)), parseFloat((r * Math.sin(t) + p[1]).toFixed(4))];
const diamPoints = (p1, p2) => p1.map((e, i) => parseFloat((e + (p2[i] - e) / 2).toFixed(4)));
function getCircles(...args) {
const [p1, p2, s] = args;
const solve = solveF(p1, s);
const halfDist = hDist(p1, p2);
let msg = [];
switch (Math.sign(s - halfDist)) {
case 0:
msg = s ? diamPoints(p1, p2) :
'Radius Zero';
break;
case 1:
if (!halfDist) {
msg = 'Coincident point. Infinite solutions';
}
else {
const theta = pAng(p1, p2);
const theta2 = Math.acos(halfDist / s);
[1, -1].map(e => solve(theta + e * theta2)).forEach(
e => msg.push(e));
}
break;
case -1:
msg = 'No intersection. Points further apart than circle diameter';
break;
default:
msg = 'Reached the default';
}
return msg;
}