Question

我有以下方法。方法rnd在两个边界之间返回一个随机整数：

  /* Create next batch of 55 random numbers */
  void advance_random (){
int j1;
double new_random;
for(j1=0; j1<24; j1++){
  new_random = oldrand[j1]-oldrand[j1+31];
  if(new_random<0.0){
    new_random = new_random+1.0;
  }
  oldrand[j1] = new_random;
}
for(j1=24; j1<55; j1++){
  new_random = oldrand[j1]-oldrand[j1-24];
  if(new_random<0.0){
    new_random = new_random+1.0;
  }
  oldrand[j1] = new_random;
}
 } //advance_ramdom

  /* Fetch a single random number between 0.0 and 1.0 */
  double randomperc(){
jrand++;
if(jrand>=55){
  jrand = 1;
  advance_random();
}
return((double)oldrand[jrand]);
  } //randomPerc

  /* Fetch a single random integer between low and high including the bounds */
  synchronized int rnd (int low, int high){
int res;
if (low >= high){
  res = low;
} else {
  res = low + (int)(randomperc()*(high-low+1));
  if (res > high){
    res = high;
  }
}
return (res);
  } // rnd

如何修改此值以使返回的数字mod2 = 0？

谢谢

Answer 1

如果您可以获得[a, b]范围内的随机数，那么您只需获取[(a+1)/2, b/2]范围内的随机数并将其乘以2即可获得范围内的随机偶数{ {1}}

Answer 2

使用bit-mask强制最低有效位为零：

x = x & ~1;

Answer 3

将你从代码中得到的结果乘以两个结尾 - 仍然是随机的，然后将其除以2！

Answer 4

如何使用：

return res & ~1;

Answer 5

在Java 1.7或更高版本中，我会使用ThreadLocalRandom：

protected override void OnModelCreating(ModelBuilder modelBuilder)
    {
        // Get rid of Cascading Circular error on ModelBuilding
        foreach (var relationShip in modelBuilder.Model.GetEntityTypes().SelectMany(e => e.GetForeignKeys()))
        {
            relationShip.DeleteBehavior = DeleteBehavior.Restrict;
        }

        #region Many-to-Many Employees Cases
        modelBuilder.Entity<EmployeeCase>()
            .HasKey(ec => new { ec.EmployeeId, ec.CaseId });

        modelBuilder.Entity<EmployeeCase>()
            .HasOne(ec => ec.Employee)
            .WithMany(e => e.EmployeeCases)
            .HasForeignKey(ec => ec.EmployeeId);

        modelBuilder.Entity<EmployeeCase>()
            .HasOne(ec => ec.Case)
            .WithMany(c => c.EmployeeCases)
            .HasForeignKey(ec => ec.CaseId);
        #endregion

        base.OnModelCreating(modelBuilder);
    }

使用ThreadLocalRandom的原因解释为here。另请注意，我们对ThreadLocalRandom.nextInt（）的输入+1的原因是为了确保max包含在范围内。

返回一个随机的偶数

5 个答案: