124 lines
2.9 KiB
Markdown
124 lines
2.9 KiB
Markdown
---
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id: 5900f39e1000cf542c50feb1
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title: 'Problem 50: Consecutive prime sum'
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challengeType: 5
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forumTopicId: 302161
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dashedName: problem-50-consecutive-prime-sum
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---
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# --description--
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The prime 41, can be written as the sum of six consecutive primes:
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<div style='text-align: center;'>41 = 2 + 3 + 5 + 7 + 11 + 13</div>
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This is the longest sum of consecutive primes that adds to a prime below one-hundred.
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The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
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Which prime, below one-million, can be written as the sum of the most consecutive primes?
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# --hints--
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`consecutivePrimeSum(1000)` should return a number.
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```js
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assert(typeof consecutivePrimeSum(1000) === 'number');
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```
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`consecutivePrimeSum(1000)` should return 953.
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```js
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assert.strictEqual(consecutivePrimeSum(1000), 953);
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```
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`consecutivePrimeSum(1000000)` should return 997651.
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```js
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assert.strictEqual(consecutivePrimeSum(1000000), 997651);
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```
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# --seed--
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## --seed-contents--
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```js
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function consecutivePrimeSum(limit) {
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return true;
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}
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consecutivePrimeSum(1000000);
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```
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# --solutions--
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```js
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// Initalize prime number list with sieve
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const NUM_PRIMES = 1000000;
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const PRIMES = [2];
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const PRIME_SIEVE = Array(Math.floor((NUM_PRIMES-1)/2)).fill(true);
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(function initPrimes(num) {
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const upper = Math.floor((num - 1) / 2);
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const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2);
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for (let i = 0; i <= sqrtUpper; i++) {
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if (PRIME_SIEVE[i]) {
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// Mark value in PRIMES array
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const prime = 2 * i + 3;
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PRIMES.push(prime);
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// Mark all multiples of this number as false (not prime)
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const primeSqaredIndex = 2 * i ** 2 + 6 * i + 3;
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for (let j = primeSqaredIndex; j < upper; j += prime)
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PRIME_SIEVE[j] = false;
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}
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}
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for (let i = sqrtUpper + 1; i < upper; i++) {
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if (PRIME_SIEVE[i])
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PRIMES.push(2 * i + 3);
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}
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})(NUM_PRIMES);
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function isPrime(num) {
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if (num === 2)
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return true;
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else if (num % 2 === 0)
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return false
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else
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return PRIME_SIEVE[(num - 3) / 2];
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}
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function consecutivePrimeSum(limit) {
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// Initalize for longest sum < 100
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let bestPrime = 41;
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let bestI = 0;
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let bestJ = 5;
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// Find longest sum < limit
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let sumOfCurrRange = 41;
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let i = 0, j = 5;
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// -- Loop while current some starting at i is < limit
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while (sumOfCurrRange < limit) {
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let currSum = sumOfCurrRange;
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// -- Loop while pushing j towards end of PRIMES list
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// keeping sum under limit
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while (currSum < limit) {
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if (isPrime(currSum)) {
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bestPrime = sumOfCurrRange = currSum;
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bestI = i;
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bestJ = j;
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}
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// -- Increment inner loop
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j++;
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currSum += PRIMES[j];
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}
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// -- Increment outer loop
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i++;
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j = i + (bestJ - bestI);
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sumOfCurrRange -= PRIMES[i - 1];
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sumOfCurrRange += PRIMES[j];
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}
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// Return
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return bestPrime;
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}
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```
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