Question

我需要计算小于或等于某个N的素数，这是素数计数函数或PI函数。我有这个，但效果太慢了：

function PI(x) {
    var primes = 4;
    for (var i = 3; i <= x; i += 2) {
        if (i % 3 === 0 || i % 5 === 0 || i % 7 === 0) continue;
        var r = ~~Math.sqrt(i), p = true;
        for (var j = 2; j <= r; j++) {
            if (i % j === 0) {
                p = false;
                break;
            }
        }
        if (p)
            primes++;
    }
    return primes;
}

计算PI(1000000)需要将近一秒钟，计算PI(10000000)对我来说只需要20秒。那太慢了。有没有更快的方法呢？

Answer 1

聚会晚了一点，但这在O（n ^ 2/3）的时间和空间上有效。

function eratosthenesWithPi(n) {
  let array = [], upperLimit = Math.sqrt(n), output = [];
  let pi = [0, 0]

  for (let i = 0; i < n; i++) {
    array.push(true);
  }

  for (let i = 2; i <= upperLimit; i++) {
    if (array[i]) {
      for (var j = i * i; j < n; j += i) {
        array[j] = false;
      }
    }
  }

  let cnt = 0

  for (let i = 2; i < n; i++) {
    if (array[i]) {
      output.push(i);
      cnt++
    }

    pi.push(cnt)
  }

  return {primes: new Uint32Array(output), pi: new Uint32Array(pi)}
}

const phiMemo = []
let primes = []

function Phi(m, b) {
  if (b === 0)
    return m
  if (m === 0)
    return 0

  if (m >= 800) {
    return Phi(m, b - 1) - Phi(Math.floor(m / primes[b - 1]), b - 1)
  }

  let t = b * 800 + m

  if (!phiMemo[t]) {
    phiMemo[t] = Phi(m, b - 1) - Phi(Math.floor(m / primes[b - 1]), b - 1)
  }

  return phiMemo[t]
}

const smallValues = [0, 0, 1, 2, 2, 3]
let piValues

function primeCountingFunction(x) {
  if (x < 6)
    return smallValues[x]

  let root2 = Math.floor(Math.sqrt(x))
  let root3 = Math.floor(x ** (1/3))

  let top = Math.floor(x / root3) + 1

  if (root2 + 1 >= primes.length) {
    let res = eratosthenesWithPi(top + 2)

    primes = res.primes
    piValues = res.pi
  }

  let a = piValues[root3 + 1], b = piValues[root2 + 1]

  let sum = 0

  for (let i = a; i < b; ++i) {
    let p = primes[i]

    sum += piValues[Math.floor(x / p)] - piValues[p] + 1
  }

  let phi = Phi(x, a)

  return phi + a - 1 - sum
}

console.log(primeCountingFunction(1e8))

在JSFiddle上尝试：https://jsfiddle.net/vo0g274f/1/

对我来说，这大约需要31毫秒。我正在开发一种更节省空间的方法atm，完成后将其发布在这里。

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