freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md

id, title, challengeType, videoUrl, dashedName

id	title	challengeType	videoUrl	dashedName
5900f39e1000cf542c50feb1	问题50：连续的总和	5		problem-50-consecutive-prime-sum

--description--

素数41可以写成六个连续素数的总和：41 = 2 + 3 + 5 + 7 + 11 + 13这是连续素数的最长和，它加到低于一百的素数。连续素数低于1000的连续素数加上一个素数，包含21个项，等于953.哪个素数低于一百万，可以写成最连续素数的总和？

--hints--

consecutivePrimeSum(1000)应该返回953。

assert.strictEqual(consecutivePrimeSum(1000), 953);

consecutivePrimeSum(1000000)应该返回997651。

assert.strictEqual(consecutivePrimeSum(1000000), 997651);

--seed--

--seed-contents--

function consecutivePrimeSum(limit) {

  return true;
}

consecutivePrimeSum(1000000);

--solutions--

function consecutivePrimeSum(limit) {
  function isPrime(num) {
    if (num < 2) {
      return false;
    } else if (num === 2) {
      return true;
    }
    const sqrtOfNum = Math.floor(num ** 0.5);
    for (let i = 2; i <= sqrtOfNum + 1; i++) {
      if (num % i === 0) {
        return false;
      }
    }
    return true;
  }
  function getPrimes(limit) {
    const primes = [];
    for (let i = 0; i <= limit; i++) {
      if (isPrime(i)) primes.push(i);
    }
    return primes;
  }

  const primes = getPrimes(limit);
  let primeSum = [...primes];
  primeSum.reduce((acc, n, i) => {
    primeSum[i] += acc;
    return acc += n;
  }, 0);

  for (let j = primeSum.length - 1; j >= 0; j--) {
    for (let i = 0; i < j; i++) {
      const sum = primeSum[j] - primeSum[i];
      if (sum > limit) break;
      if (isPrime(sum) && primes.indexOf(sum) > -1) return sum;
    }
  }
}