/ ************************************************** ************
* This program uses the granted method for multiplication
* Seeking 1011 * 1101
* CODER: Nine Day Shenlong
* operation result:
* H: / java / algorithm / big integer ~ 1> javac mult.java
*
* H: / java / algorithm / big integer ~ 1> java multi
* 11x13 Results: 143
*
* H: / java / algorithm / big integer ~ 1>
*********************************************************** ************ /
Import java.lang. *;
Class Mult {
Public int x, y; // This topic requires two numbers
/ / Initialize two numbers required in this question
MULT () {
X = 11; // binary 1011
Y = 13; // binary 1101
}
// ******************************************************** **
// Function name getValue, this function is the main function of this algorithm
// Reverse value is the product of the specified number xy
/ / According to formula XY = a * c * 2
N ((A-B) (D-C) AC BD) * 2
N / 2 BD
// A total of 3 multiplication, 6 plus method, 2 shift
// ******************************************************** **
Public Int GetValue (int A, int B, int C, int D) {
INT N = 4; // Number of scores
Return (A * C * (2 << (N-1))
((A-b) * (D-C) A * C B * D) *
(2 << (n / 2-1))
B * D);
}
Public static void main (String [] args) {
INT n = 2;
Mult mu = new mult ();
INT A, B, C, D;
// Decompose the original multiplier
A = MU.X >> N;
B = MU.X & 3;
C = MU.Y >> N;
D = MU.Y & 3;
int RET = MU.GETVALUE (A, B, C, D);
// Output Result
System.out.println (" MU.X " X " MU.Y " results are: " RET);
}
}
