[自娱自乐]模拟像素打印和位练习

参考文章：

http://blog.csdn.net/sxhelijian/article/details/17167291

原理：

1、按位打印字符，并用32位整型来压缩.（十六进制表示）

2、位操作： 0 & x = 0, 1 | x = 1.  位状态的反转

3、轴对称的利用

#include <iostream>
using namespace std;

//digit table
int heart[21] = {
	0x420, 0x4004, 0x10000, 0x40000,
	0x100000, 0x200000, 0x600000, 0x200000, 0x100000,
	0x40000, 0x8000, 0x800, 0x100, 0x10,
	0x4, 0x1
};

//reverse bit status
int bitReverse(int n)
{
	int sum = 0, tempBit;
	for (int i = 0; i < 32; ++i)
	{
		tempBit = n % 2;
		n /= 2;
		sum <<= 1;
		if (tempBit == 0)
		{
			sum &= 0xfffffffe;
		}
		else 
		{
			sum |= 0x1;
		}
	}
	return sum;
}

//print character according to given value
void printPixel(int n)
{
	if (0 == n)
		cout << " ";
	else 
		cout << "$";
}

//obtain bit status, and print
void print(int num)
{
	for (int j = 0; j < 32; ++j)
	{
		printPixel(num % 2);
		num /= 2;
	}
}

int main(int argc, char const *argv[])
{
	for (int i = 1; i <= 21; ++i)	//Row control
	{
		//axial symmetry
		int temp = bitReverse(heart[i-1]);
		print(temp);
		
		switch(i)
		{
			case 4:
			case 3:
				cout << "$";
			break;
		}	//axial operation

		print(heart[i - 1]);
		cout << endl;
	}
	return 0;
}

[自娱自乐]模拟像素打印和位练习

原文：http://blog.csdn.net/tbz888/article/details/19488729