1.2 Question of Chinese chess masters further discussion and expansion: how to use one variable to implement N-repeating? [Chinese chess]

[Question]

Assume that only two pieces of chess are left in Chinese chess. Chinese people all know the basic rules: the handsome guy can only move up or down the left and right sides, but cannot move in oblique directions. In this case, all possible positions will be handsome.Only one byte storage variable is required in the code..

[Analysis]

Three solutions:

1)Bitwise operationOne byte is used to store and read two variables.

2) UseBit domainSeveral different objects are represented by a byte binary field. For example

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Struct { Unsigned char A: 4; Unsigned char B: 4; } I;

3) use one variable to express a 2-cycle. This solution will be discussed later. (THINKING: how to use one variable to implement N-repeating ?)

Bitwise operation]

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/* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/6/24 */ # Include <stdio. h> # Define half_bits_length 4 # Define fullmask 255 # Define lmask (fullmask # Define rmask (fullmask> half_bits_length) # Define rset (B, n) (B = (B & lmask) | (n )) # Define lset (B, n) (B = (B & rmask) | (n) # Define rget (B) (B & rmask) # Define lget (B) (B & lmask)> half_bits_length) # Define gridw 3 Void solution1 () { Unsigned char B; For (lset (B, 1); lget (B) <= gridw * gridw; lset (B, (lget (B) + 1 ))) { For (rset (B, 1); rget (B) <= gridw * gridw; rset (B, (rget (B) + 1 )) { If (lget (B) % gridw! = Rget (B) % gridw) { Printf ("A = % d, B = % d \ n", lget (B), rget (B )); } } } }

【Bit domain]

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/* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/6/24 */ Struct { Unsigned char A: 4; Unsigned char B: 4; } I; Void solution2 () { For (I. A = 1; I. A <= 9; I. A ++) For (I. B = 1; I. B <= 9; I. B ++) If (I. A % 3! = I. B % 3) // column not equal Printf ("% d, % d \ n", I. A, I. B ); }

[Single Variable]

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/* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/6/24 */ Void solution3 () { Unsigned char I = 81; While (I --) { // I = 9 * A + B, A = I/9, B = I % 9 If (I/9% 3 = I %9% 3) Continue; Printf ("% d, % d \ n", I/9, I % 9 ); } }

We need to verify the 9*9 = 81 positional relationships between the "Jiang" and "Shuai" in our 3*3 lattice, which is also the origin of I = 81. In addition, we need to understand the meanings of I/9 and I % 9. We know that the integer I can be composed of two parts: I = (I/9) * 9 + I % 9. We note that in the process of I from 81 to 0, the change of I % 9 is equivalent to the inner loop, and the change of I/9 is relative to the outer loop.

[Expansion]

How to use one variable to implement N-repeating?

Let's take a look at a simple example. One Variable achieves a 2-repeating loop.

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/* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/6/24 */ Void loopwith2variables () { Unsigned char I, J; For (I = 0; I <5; I ++) For (j = 0; j <4; j ++) Printf ("% d, % d", I, j ); } Void loopwith1variable () { Unsigned char val = 4*5; While (Val --) { Printf ("% d, % d", (Val/4) % 5, Val % 4 ); } }

[Summary]

For a * B = I, use the following formula:

Loop1 = I % B;

Loop2 = (I/B) %

Here, loop1 is the inner loop, and loop2 is the outer loop.

Then we can obtain the n-weight formula. Suppose an * a (n-1) * ...... * A3 * A2 * a1 = N

Loop1 = n % A1

Loop2 = (N/(A1) % A2

Loop3 = (N/(a1a2) % A3

.....

Loopn = (N/(a1a2... A (n-1) %

[Reference]

Http://blog.csdn.net/kabini/article/details/2256421

Http://blog.csdn.net/silenceburn/article/details/6133222

Http://blog.csdn.net/zhongkeli/article/details/8779168

Http://www.cnblogs.com/python27/archive/2012/04/10/2441114.html

[Link to this article]

Http://www.cnblogs.com/hellogiser/p/chinese-chess.html