* 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>
1.5 KiB
1.5 KiB
id, title, challengeType, forumTopicId, dashedName
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
a3bfc1673c0526e06d3ac698 | 质数求和 | 5 | 16085 | sum-all-primes |
--description--
质数是大于 1 且仅可以被 1 和自己整除的数。比如,2 就是一个质数,因为它只可以被 1 和 2(它本身)整除。相反,4 不是质数,因为它可以被 1, 2 和 4 整除。
请完成 sumPrimes
方法,使其返回小于等于传入参数数字的所有质数之和。
--hints--
sumPrimes(10)
应返回一个数字。
assert.deepEqual(typeof sumPrimes(10), 'number');
sumPrimes(10)
应返回 17。
assert.deepEqual(sumPrimes(10), 17);
sumPrimes(977)
应返回 73156。
assert.deepEqual(sumPrimes(977), 73156);
--seed--
--seed-contents--
function sumPrimes(num) {
return num;
}
sumPrimes(10);
--solutions--
function eratosthenesArray(n) {
var primes = [];
if (n > 2) {
var half = n>>1;
var sieve = Array(half);
for (var i = 1, limit = Math.sqrt(n)>>1; i <= limit; i++) {
if (!sieve[i]) {
for (var step = 2*i+1, j = (step*step)>>1; j < half; j+=step) {
sieve[j] = true;
}
}
}
primes.push(2);
for (var p = 1; p < half; p++) {
if (!sieve[p]) primes.push(2*p+1);
}
}
return primes;
}
function sumPrimes(num) {
return eratosthenesArray(num+1).reduce(function(a,b) {return a+b;}, 0);
}
sumPrimes(10);