El maximo valor de enteros en java y el 0x3f3f3f3f

stackoverflow.com/questions/5771520/why-is-the-maximum-value-of-an-unsigned-n-bit-integer-2n-1-and-not-2n

porque teniendo un valor entero de 32 bits su maximo valor es 2^31-1 y no 2^31-1 
Respuesta

@HardikThaker - Yes, it does. (Actually, in Java, Integer.MAX_VALUE--for 32 bit integers--is 2^31 - 1 andLong.MAX_VALUE--for 64 bit integers--is 2^63 - 1 because the sign bit is reserved.) – Ted Hopp Sep 29 '13 at 0:03 


ADEMAS que es el  0x3f3f3f3f

0x3f3f3f3f; // Numero muy grande para que no afecte al minHeap.

 //memset(m,0x3f,sizeof(m));

0x3f3f3f3f + 0x3f3f3f3f doesn't overflow int32

http://prog3.com/sbdm/blog/u013163567/article/details/19837805
http://ideone.com/cqFDL5
http://stackoverflow.com/questions/18429021/why-is-infinity-0x3f3f3f3f

Operador modulo con numeros negativos

1down vote

The Quickest and easiest way to find the mod of a negative number is by using the below property if a = (b) mod c then a = (c*k + b) mod c (where k = 1,2,3.......) It simply says that the value of a is unchanged when we add a multiple of c to b Example a = (10) mod 3 we all know that a = 1 Now a = (3*1 + 10) mod 3 - a is still = 1 a = (3*2 + 10) mod 3 - a is still = 1 a = (3*3 + 10) mod 3 - a is still = 1 a = (3*4 + 10) mod 3 - a is still = 1 So adding any multiple of 3 (> 0) to 10 does not effect the value of a Now we use this to our advantage in finding mod of negative numbers Example a = (-10) mod 3 Now i add 12 to 10 as 12 is a multiple of 3 and hence the value of a will remain unchanged so a = (3*4 – 10) mod 3 = 2 mod 3 = 2 easy isnt it? Another example a = (-340) mod 60 So a = (60*6 – 340) mod 60 = (360-340) mod 60 = 20 mod 60 = 20

**************
To find −bmodN$-b\mathrm{mod}N$, just keep adding N$N$ to −b$-b$ until the number is between 0 and N$N$.

As an example, N=13,b=−27$N=13,b=-27$. Add 13 to -27, you get -14, again you get -1, and again you get 12.

So, −27mod13=12$-27\mathrm{mod}13=12$.

A bit more generally, you might want to realize that amodN=a+kNmodN$a\mathrm{mod}N=a+kN\mathrm{mod}N$ for any k∈N$k\in \mathbb{N}$. That should help with your first question.

***************

A encontrar −bmodN$-b\mathrm{mod}N$, sólo mantenga la adición de N$N$ a b$b$ hasta que el número está entre 0 y N$N$.

Como un ejemplo, N=13,b=−27$N=13,b=-27$. Añadir 13 a -27, consigue -14, de nuevo usted obtiene -1, y de nuevo consigue 12.

Así, −27mod13=12$-27\mathrm{mod}13=12$.

Un poco más generalmente, usted podría querer darse cuenta de que amodN=a+kNmodN$a\mathrm{mod}N=a+kN\mathrm{mod}N$ para k∈N$k\in \mathbb{N}$. Eso debería ayudar con tu primera pregunta.
***************

Based on the C99 Specification: a = (a / b) * b + a % b

We can write a function to calculate (a % b) = a - (a / b) * b!

int remainder(int a, int b)
{
    return a - (a / b) * b;
}

For modulo operation, we can have the following function:

int mod(int a, int b)
{
    int r = a % b;
    return r < 0 ? r + b : r;
}

My conclusion is (a % b) in C is a remainder operator and NOT modulo operator.