Ainsi, vous pouvez ajouter des données supplémentaires au schéma généré automatiquement.

Par exemple, vous pouvez décider de lire et de valider la requête avec votre propre code, sans utiliser les fonctionnalités automatiques de validation proposée par Pydantic, mais vous pouvez toujours définir la requête dans le schéma OpenAPI.

size is automatic based on the number of drives available, let's say for example if there are 32 servers and 32 drives which is a total of 1024 drives. In this scenario 16 becomes the erasure set size. This is decided based on the greatest common divisor (GCD) of acceptable erasure set sizes ranging from *4 to 16*.

- *If total drives has many common divisors the algorithm chooses the minimum amounts of erasure sets possible for a erasure set size of any N*.  In the example with 1024 drives...

      // non-negative int, we can do long-arithmetic on index * (dataset.length - 1) / scale to get
      // a rounded ratio and a remainder which can be expressed as ints, without risk of overflow:
      int quotient = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
      int remainder = (int) (numerator - (long) quotient * scale);
      selectInPlace(quotient, dataset, 0, dataset.length - 1);
      if (remainder == 0) {

func DivModInt(q, r, x, y *Int) {
	q.doinit()
	r.doinit()
	x.doinit()
	y.doinit()
	C.mpz_tdiv_qr(&q.i[0], &r.i[0], &x.i[0], &y.i[0])
}

// GcdInt sets d to the greatest common divisor of a and b,
// which must be positive numbers.
// If x and y are not nil, GcdInt sets x and y such that d = a*x + b*y.
// If either a or b is not positive, GcdInt sets d = x = y = 0.
func GcdInt(d, x, y, a, b *Int) {

	DIVDUVCC R3, R4, R5             // 7ca41f93
	DIVWUVCC R3, R4, R5             // 7ca41f97
	DIVWUV   R3, R4, R5             // 7ca41f96
	DIVDE R3, R4, R5                // 7ca41b52
	DIVDECC R3, R4, R5              // 7ca41b53
	DIVDEU R3, R4, R5               // 7ca41b12
	DIVDEUCC R3, R4, R5             // 7ca41b13

	REM R3, R4, R5                  // 7fe41bd67fff19d67cbf2050

  public void testBitSize() {
    double fpp = 0.03;
    for (int i = 1; i < 10000; i++) {
      long numBits = BloomFilter.optimalNumOfBits(i, fpp);
      int arraySize = Ints.checkedCast(LongMath.divide(numBits, 64, RoundingMode.CEILING));
      assertEquals(
          arraySize * Long.SIZE,
          BloomFilter.create(Funnels.unencodedCharsFunnel(), i, fpp).bitSize());
    }
  }

  public void testBitSize() {
    double fpp = 0.03;
    for (int i = 1; i < 10000; i++) {
      long numBits = BloomFilter.optimalNumOfBits(i, fpp);
      int arraySize = Ints.checkedCast(LongMath.divide(numBits, 64, RoundingMode.CEILING));
      assertEquals(
          arraySize * Long.SIZE,
          BloomFilter.create(Funnels.unencodedCharsFunnel(), i, fpp).bitSize());
    }
  }

	VMULHSUVV	V1, V2, V3			// d7a1209a
	VMULHSUVV	V1, V2, V0, V3			// d7a12098
	VMULHSUVX	X10, V2, V3			// d761259a
	VMULHSUVX	X10, V2, V0, V3			// d7612598

	// 31.11.11: Vector Integer Divide Instructions
	VDIVUVV		V1, V2, V3			// d7a12082
	VDIVUVV		V1, V2, V0, V3			// d7a12080
	VDIVUVX		X10, V2, V3			// d7612582
	VDIVUVX		X10, V2, V0, V3			// d7612580
	VDIVVV		V1, V2, V3			// d7a12086

	oldCache    dataUsageCache
	newCache    dataUsageCache
	updateCache dataUsageCache

	dataUsageScannerDebug bool
	healObjectSelect      uint32 // Do a heal check on an object once every n cycles. Must divide into healFolderInclude
	scanMode              madmin.HealScanMode

	weSleep    func() bool
	shouldHeal func() bool

	disks       []StorageAPI
	disksQuorum int

Levaria o mesmo tempo para finalizar com ou sem turnos (concorrência) e você teria feito o mesmo tanto de trabalho.

Mas nesse caso, se você trouxesse os 8 ex-caixas / cozinheiros / agora-faxineiros, e cada um deles (mais você) pudessem dividir a casa para limpá-la, vocês fariam toda a limpeza em **paralelo**, com a ajuda extra, e terminariam muito mais cedo.

Nesse cenário, cada um dos faxineiros (incluindo você) poderia ser um processador, fazendo a sua parte do trabalho.