ee1e8abd87
* 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>
1.3 KiB
1.3 KiB
id, title, challengeType, videoUrl, dashedName
| id | title | challengeType | videoUrl | dashedName |
|---|---|---|---|---|
| 5900f3721000cf542c50fe85 | 问题6：求和方差 | 5 | problem-6-sum-square-difference |
--description--
前十个自然数的平方和是，
1
2
- 2
2
- ... + 10
2
= 385
前十个自然数之和的平方是，
（1 + 2 + ... + 10）
2
= 55
2
= 3025
因此，前十个自然数的平方和与和的平方之间的差值为3025 - 385 = 2640.求出前n自然数的平方和与总和的平方之间的差值。
--hints--
sumSquareDifference(10)应该返回2640。
assert.strictEqual(sumSquareDifference(10), 2640);
sumSquareDifference(20)应该返回41230。
assert.strictEqual(sumSquareDifference(20), 41230);
sumSquareDifference(100)应该返回25164150。
assert.strictEqual(sumSquareDifference(100), 25164150);
--seed--
--seed-contents--
function sumSquareDifference(n) {
return true;
}
sumSquareDifference(100);
--solutions--
const sumSquareDifference = (number)=>{
let squareOfSum = Math.pow(sumOfArithmeticSeries(1,1,number),2);
let sumOfSquare = sumOfSquareOfNumbers(number);
return squareOfSum - sumOfSquare;
}
function sumOfArithmeticSeries(a,d,n){
return (n/2)*(2*a+(n-1)*d);
}
function sumOfSquareOfNumbers(n){
return (n*(n+1)*(2*n+1))/6;
}