Author Archives: neohope

#include "stdafx.h"
#include <nmmintrin.h>

/*
 *通过移位计算1的个数
 *每一位都要判断和处理
 */
int BitCount(unsigned int n)
{
	unsigned int c = 0;
	while (n >0)
	{
		if ((n & 1) == 1)++c;
		n >>= 1;
	}
	return c;
}

/*
 *通过减法及位运算，保证每次至少消除一个1
 *只处理1的位，0的位不处理
 */
int BitCountWithMinus(unsigned int n)
{
	unsigned int c = 0;
	for (c = 0; n; ++c)
	{
		n &= (n - 1);
	}
	return c;
}

/*
*将32位数字，截为4个8位数
*通过查表，得到每个8位数中1的个数，然后求和
*/
int BitCountLUT8(unsigned int n)
{
	unsigned char BitsSetTable256[256] = { 0 };

	for (int i = 0; i <256; i++)
	{
		BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];
	}

	unsigned int c = 0;

	unsigned char* p = (unsigned char*)&n;

	c = BitsSetTable256[p[0]] +
		BitsSetTable256[p[1]] +
		BitsSetTable256[p[2]] +
		BitsSetTable256[p[3]];

	return c;
}

/*
*将32位数字，截为4个8位数
*通过查表，得到每个8位数中1的个数，然后求和
*LUT表已经计算好
*/
int BitCountLUT8Static(unsigned int n)
{
	unsigned int table[256] =
	{
		0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,
	};

	return table[n & 0xff] +
		table[(n >> 8) & 0xff] +
		table[(n >> 16) & 0xff] +
		table[(n >> 24) & 0xff];
}

/*
*将32位数字，截为8个4位数
*通过查表，得到每个4位数中1的个数，然后求和
*LUT表已经计算好
*/
int BitCountLUT4Static(unsigned int n)
{
	unsigned int table[16] =
	{
		0, 1, 1, 2,
		1, 2, 2, 3,
		1, 2, 2, 3,
		2, 3, 3, 4
	};

	unsigned int count = 0;
	while (n)
	{
		count += table[n & 0xf];
		n >>= 4;
	}
	return count;
}

/*
*将32位数字，相邻的2位求和，然后相邻的4位求和
*然后相邻的8位、16位、32位求和
*/
int BitCountParallel(unsigned int n)
{
	n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
	n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
	n = (n & 0x0f0f0f0f) + ((n >> 4) & 0x0f0f0f0f);
	n = (n & 0x00ff00ff) + ((n >> 8) & 0x00ff00ff);
	n = (n & 0x0000ffff) + ((n >> 16) & 0x0000ffff);

	return n;
}

/*
*第一行，计算每三位的1的个数（其实11*3是33位，但最高一位可以假设为0，所以没问题）
*第二行，实际上是先计算了相邻6位中1的个数（不会产生进位，结果最多占到3位），并将前三位置为0
*然后通过取模，相当于将每6位数字做了加法
*/
int BitCountMagic(unsigned int n)
{
	unsigned int tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111);
	return ((tmp + (tmp >> 3)) & 030707070707) % 63;
}

unsigned int n = 127;
unsigned int bitCount = _mm_popcnt_u32(n);


/*
*将8位数字，转换为MY_UNSIGHED_CHAR结构体，然后求和
*/
struct MY_UNSIGHED_CHAR
{
	unsigned a : 1;
	unsigned b : 1;
	unsigned c : 1;
	unsigned d : 1;
	unsigned e : 1;
	unsigned f : 1;
	unsigned g : 1;
	unsigned h : 1;
};

long BitCountStuct(unsigned char b)
{
	struct MY_UNSIGHED_CHAR *by = (struct MY_UNSIGHED_CHAR*)&b;
	return (by->a + by->b + by->c + by->d + by->e + by->f + by->g + by->h);
}

int _tmain(int argc, _TCHAR* argv[])
{
	printf("There are %d 1 in 133\n", BitCount(133));
	printf("There are %d 1 in 133\n", BitCountWithMinus(133));
	printf("There are %d 1 in 133\n", BitCountLUT8(133));
	printf("There are %d 1 in 133\n", BitCountLUT8Static(133));
	printf("There are %d 1 in 133\n", BitCountLUT4Static(133));
	printf("There are %d 1 in 133\n", BitCountParallel(133));
	printf("There are %d 1 in 133\n", BitCountMagic(133));
	printf("There are %d 1 in 133\n", BitCountStuct(133));
	return 0;
}

        //加法
	public static int add(int a, int b)  
	{  
	    int ans;  
	    while(b!=0)  
	    {
	        ans = a^b;
	        b = ((a&b)<<1);
	        a = ans;  
	    }  
	    return a;  
	}  
	
	//减法
	public static int sub(int a, int b)
	{  
	    return add(a, -b);  
	}
	
	//正数乘法
	private static int posMultiply(int a,int b)  
	{  
	    int ans = 0;  
	    while(b>0)  
	    {  
	        if((b&0x1)==1)ans = add(ans, a);  
	        a = (a<<1);  
	        b = (b>>1);  
	    }  
	    return ans;  
	}  
	  
	//乘法
	public static int multiply(int a,int b)  
	{  
	    if(a==0||b==0)return 0;  
	    if(a>0 && b>0)  
	        return posMultiply(a, b);  
	    if(a<0)  
	    {  
	        if(b<0)  
	        {  
	            return posMultiply(-a, -b);  
	        } 
	        else
	        {
	        	return -posMultiply(-a, b ); 
	        }
	    }
	    else
	    {
	    	return -posMultiply(a, -b);
	    }
	}  
	  
	//正数除法
	private static int posDiv(int x,int y)  
	{  
	    int ans=0;  
	    for(int i=31;i>=0;i--)  
	    {
	        if((x>>i)>=y)  
	        {  
	            ans+=(1<<i);  
	            x-=(y<<i);  
	        }  
	    }  
	    return ans;  
	}  
	  
	//除法
	public static int div( int a, int b )  
	{
		assert(b!=0);
	    if(a==0)return 0;

	    if(a>0)  
	    {  
	        if(b>0)
	        {
	            return posDiv(a,b); 
	        }
	        else
	        {
	        	return -posDiv( a,-b);  
	        }
	    }  
	    else
	    {
	    	if(b>0)
	    	{
		        return -posDiv(a, b);  
	    	}
	    	else
	    	{
	    		return -posDiv(-a, -b);  
	    	}
	    }
	    
	}   
	
	//求负数
	public static int negtive(int a)
	{  
	    return add(~a, 1);  
	}
	  
	//比较正数大小
	private static boolean isbigerPos( int a, int b )   
	{
	    int c = 1;  
	    b = (a^b);  
	    if(b==0)return false;
	    
	    while(b>0)  
	    {  
	    	b>>=1;
	        c <<= 1;  
	    }  
	    return (c&a)==0;  
	}   
	  
	//比较大小   
	public static boolean isbiger( int a, int b )   
	{
	    if(a<0)  
	    {  
	        if(b<0)  
	        {  
	            return isbigerPos( negtive(b), negtive(a) );  
	        }
	        else
	        {
	        	return false;  
	        }
	    }
	    else
	    {
		    if(b<0)
		    {
		        return true;  
		    }
		    else
		    {
		    	return isbigerPos(a, b);  
		    }
	    }

	}

	public static int divideby3(int x)  
	{  
	    int sum = 0;  
	    while(x > 3)  
	    {  
	        sum = add(x>>2 , sum);  
	        x = add(x>>2 , x&3);  
	    } 
	    if(x == 3) 
	    {
	        sum = add(sum , 1);
	    }
	    return sum;  
	}  
	
	public static void main(String[] args)
	{
		System.out.println(add(13,19));
		System.out.println(sub(13,19));
		System.out.println(multiply(13,19));
		System.out.println(div(10,5));
		System.out.println(negtive(13));
		System.out.println(isbiger(10,5));
		System.out.println(isbiger(10,11));
		
		System.out.println(divideby3(15));
		System.out.println(divideby3(14));
		System.out.println(divideby3(13));
		System.out.println(divideby3(12));
		System.out.println(divideby3(11));
	}

        //精度
	private final static double accuracy= 1e-6;

	/**
	 * 暴力求解
	 */
	public static double bruteSqrt(double x)
	{
		assert(x>=0);
		double ans=0.0;
		while (Math.abs(x - ans * ans) > accuracy)ans += accuracy;
		return ans;
	}

	/**
	 * 牛顿法求解
	 */
	public static double newtonSqrt(double x)
	{
		assert(x>=0);
		double avg = x;
		double last_avg = Double.MAX_VALUE;

		while (Math.abs(avg - last_avg) > accuracy)
		{
			last_avg = avg;
			avg = (avg + x / avg) / 2;
		}
		return avg;
	}

	/**
	 * 二分法求解
	 */
	public static double binarySqrt(double x)
	{
		assert(x>=0);

		double low = 0;
		double high = x;
		double mid = Double.MAX_VALUE;
		double last_mid = Double.MIN_VALUE;

		while (Math.abs(mid - last_mid) > accuracy)
		{
			last_mid = mid;
			mid = (low + high)/2;
			if (mid*mid>x)high = mid;
			if (mid*mid<x)low = mid;
		}
		return mid;

	}

	private final static int[] LUT =
	{ 0, 16, 22, 27, 32, 35, 39, 42, 45, 48, 50, 53, 55, 57, 59, 61, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 81, 83, 84,
			86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 103, 104, 106, 107, 108, 109, 110, 112, 113, 114, 115,
			116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 128, 128, 129, 130, 131, 132, 133, 134, 135, 136,
			137, 138, 139, 140, 141, 142, 143, 144, 144, 145, 146, 147, 148, 149, 150, 150, 151, 152, 153, 154, 155,
			155, 156, 157, 158, 159, 160, 160, 161, 162, 163, 163, 164, 165, 166, 167, 167, 168, 169, 170, 170, 171,
			172, 173, 173, 174, 175, 176, 176, 177, 178, 178, 179, 180, 181, 181, 182, 183, 183, 184, 185, 185, 186,
			187, 187, 188, 189, 189, 190, 191, 192, 192, 193, 193, 194, 195, 195, 196, 197, 197, 198, 199, 199, 200,
			201, 201, 202, 203, 203, 204, 204, 205, 206, 206, 207, 208, 208, 209, 209, 210, 211, 211, 212, 212, 213,
			214, 214, 215, 215, 216, 217, 217, 218, 218, 219, 219, 220, 221, 221, 222, 222, 223, 224, 224, 225, 225,
			226, 226, 227, 227, 228, 229, 229, 230, 230, 231, 231, 232, 232, 233, 234, 234, 235, 235, 236, 236, 237,
			237, 238, 238, 239, 240, 240, 241, 241, 242, 242, 243, 243, 244, 244, 245, 245, 246, 246, 247, 247, 248,
			248, 249, 249, 250, 250, 251, 251, 252, 252, 253, 253, 254, 254, 255 };

	/**
	 * 查表法求解
	 */
	public static int intLutSqrt(int x)
	{
		int xn;

		if (x >= 0x10000)
		{
			if (x >= 0x1000000)
			{
				if (x >= 0x10000000)
				{
					if (x >= 0x40000000)
					{
						xn = LUT[x >> 24] << 8;
					}
					else
					{
						xn = LUT[x >> 22] << 7;
					}
				}
				else
				{
					if (x >= 0x4000000)
					{
						xn = LUT[x >> 20] << 6;
					}
					else
					{
						xn = LUT[x >> 18] << 5;
					}
				}

				xn = (xn + 1 + (x / xn)) >> 1;
				xn = (xn + 1 + (x / xn)) >> 1;
				return ((xn * xn) > x) ? --xn : xn;
			}
			else
			{
				if (x >= 0x100000)
				{
					if (x >= 0x400000)
					{
						xn = LUT[x >> 16] << 4;
					}
					else
					{
						xn = LUT[x >> 14] << 3;
					}
				}
				else
				{
					if (x >= 0x40000)
					{
						xn = LUT[x >> 12] << 2;
					}
					else
					{
						xn = LUT[x >> 10] << 1;
					}
				}

				xn = (xn + 1 + (x / xn)) >> 1;

				return ((xn * xn) > x) ? --xn : xn;
			}
		}
		else
		{
			if (x >= 0x100)
			{
				if (x >= 0x1000)
				{
					if (x >= 0x4000)
					{
						xn = (LUT[x >> 8]) + 1;
					}
					else
					{
						xn = (LUT[x >> 6] >> 1) + 1;
					}
				}
				else
				{
					if (x >= 0x400)
					{
						xn = (LUT[x >> 4] >> 2) + 1;
					}
					else
					{
						xn = (LUT[x >> 2] >> 3) + 1;
					}
				}

				return ((xn * xn) > x) ? --xn : xn;
			}
			else
			{
				if (x >= 0)
				{
					return LUT[x] >> 4;
				}
			}
		}

		return -1;
	}

	/**
	 * Quake III中快速求解平方根倒数的方法
	 */
	public static float fastInvSqrt(float x)  
	{  
	     float xhalf = 0.5f*x;  
	     int f2i = Float.floatToRawIntBits(x);
	     f2i = 0x5f375a86-(f2i>>1);
	     x = Float.intBitsToFloat(f2i);
	     x = x*(1.5f-xhalf*x*x);
	     x = x*(1.5f-xhalf*x*x);
	     return x;  
	}
	
	/**
	 * Quake III中快速求解平方根方法
	 */
	public static float fastSqrt(float x) {  
	    float y=x;
	    float xhalf = 0.5f*x;
	    int f2i = Float.floatToRawIntBits(x);  
	    f2i = 0x5f3759df-(f2i>>1);  
	    x = Float.intBitsToFloat(f2i);  
	    x  = x * (1.5f-(xhalf*x*x));  
	    x  = x * (1.5f-(xhalf*x*x));  
	    return y*x;  
	}

	public static void main(String[] args)
	{
		System.out.println(bruteSqrt(3));
		System.out.println(newtonSqrt(3));
		System.out.println(binarySqrt(3));
		System.out.println(intLutSqrt(64));
		System.out.println(1/fastInvSqrt(3));
		System.out.println(fastSqrt(3));
	}