#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;
}
Category Archives: Algorithm
用位运算实现加减乘除
//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));
}
几种常见求解平方根的方法
//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));
}