tiandyoin

种群计数 (pop_count)

//高效程序的奥秘hacker's delight Henry S. Warren, Jr. 著冯速译

//第5 章位计数5.1 1 位计数

// 应用：

// Hamming 距离是矢量间取不同值的对应位的数目,即dist(x, y) = pop(x ^ y);

// 稀疏数组A 压缩后的简便访问方式，可以利用位串数组BITS 记录相应下标有定义的元素,

// 如每个有定义的A[i]，BITS 的第I 个元素是1 位。

#include "iostream"

#include "bitset"

#include "limits"

#include "time.h"

using namespace std;

#define out2(T) cout<<bitset<numeric_limits<unsigned int>::digits>(T)<<endl;

#define SIZE(T) (numeric_limits<unsigned T>::digits)

bool SHR31(int x) //逻辑右移位

{

return (unsigned int)x>>31;

}

#define HOUT(x,y,z) hex_out(x,y,z)

inline void hex(int x)

{

int w = cout.width();

cout << "0x";

cout.width(8); cout.fill('0');

cout << hex << x << " ";

cout.width(w);

}

inline void hex(unsigned __int64 x)

{

int w = cout.width();

cout << "0x";

cout.width(16); cout.fill('0');

printf("%I64x ",x);

cout.width(w);

}

void hex_out(int x, int y, int z)

{

hex(x); hex(y); hex(z); cout << endl;

}

// 循环左移位

int rotate(unsigned int x, int n)

{

return (x<<n) | (x >> (32-n));

}

// 统计x 中位1 的个数

// 首先设置每个2 位字段为原来的两个单位的和，然后，求相邻位字段的和，把结果放入相应的位字段，以此类推。

int pop(unsigned int x)

{

x = (x & 0x55555555) + ((x >> 1) & 0x55555555);

x = (x & 0x33333333) + ((x >> 2) & 0x33333333);

x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);

x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);

x = (x & 0x0000FFFF) + ((x >>16) & 0x0000FFFF);

return x;

}

// 对x 的第一个赋值基于下面这个相当巧妙的公式的前两项

// pop(x) = x - floor(x/2) - floor(x/4) - ... - floor(x/2^31);

// 设字是b3b2b1b0

// bi = floor(x/2^i) - 2*floor(x/2^(i+1))

int pop2(unsigned int x)

{

x = x - ((x >> 1) & 0x55555555);

x = (x & 0x33333333) + ((x >> 2) & 0x33333333);

x = (x + (x >> 4)) & 0x0F0F0F0F;

x = x + (x >> 8);

x = x + (x >>16);

return x & 0x00000003F;

}

// [HAK, item169]

// http://www.inwap.com/pdp10/hbaker/hakmem/hacks.html

// 在DEC PDP-10计算机上只需要10 个指令

// 它是十进制计数的计算机，mod 63 很方便

// 注意：以0 开头的数是八进制数

// 前三项萃取位字段的数目

// 举例：如每三位字段全 111B

// (1): 取出高二位对应1 的值, 有两个1 ，则取出后n 为011

// (3): 取出最高位对应1 的值，有一个1 ，则取出后n 为001

// (2),(4): x - n - n , 刚好11X - 011 - 001 为2+X ＝高二位1 的数目+X

// 而最低位X 为1 则自然加1，为0 自然加0

// (5): 合并相邻的两个三位字段成六位字段

// (6): mod 63 是怎么操作的呢？

// 假设六位字段值为a6 a5 a4 a3 a2 a1 a0

// 十进制和为a6*64^5 + a5*64^4 + ... + a0

// = a6* (63+1)^5 + a5* (63+1)^4 + ... + a0

// mod 63 刚好剩下a6 + a5 + ... + a0

// 从而把所有六位字段相加，统计1 的总和

int pop3(unsigned int x)

{

#define modu(x, y) (x % 63)

unsigned int n;

n = (x >> 1) & 033333333333;

x = x - n;

n = (n >> 1) & 033333333333;

x = x - n;

x = (x + (x >> 3)) & 030707070707;

x = modu(x, 63);

return x;

}

int pop4(unsigned int x)

{

unsigned int n;

n = (x >> 1) & 0x77777777;

x = x - n;

n = (n >> 1) & 0x77777777;

x = x - n;

n = (n >> 1) & 0x77777777;

x = x - n;

x = (x + (x >> 4)) & 0x0F0F0F0F;

x = x * 0x01010101;

return x >> 24;

}

int pop5(unsigned int x)

{

int n;

n = 0;

while (x != 0) {

n += 1;

x = x & (x - 1); // 把最右侧的1 位改成0 位

}

return n;

}

int pop6(unsigned int x) {

int i, sum;

sum = x;

for (i = 1; i <= 31; i++) {

x = rotate(x, 1);

sum += x;

}

return -sum;

}

// 只计算七位量

int pop7(unsigned int x) {

x = x * 0x02040810;

x = x & 0x11111111;

x = x * 0x11111111;

x = x >> 28;

return x;

}

// 只计算八位量

int pop8(unsigned int x) {

x = x * 0x08040201;

x = x >> 3;

x = x & 0x11111111;

x = x * 0x11111111;

x = x >> 28;

return x;

}

int pop9(unsigned int x) {

int sum;

sum = x;

while (x != 0) {

x = x >> 1;

sum = sum - x;

}

return sum;

}

// table 存放0 到255 的x 的所有pop(x) 值

int pop10(unsigned int x) {

static char 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,

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,

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[x & 0xFF] +

table[(x >> 8) & 0xFF] +

table[(x >>16) & 0xFF] +

table[(x >>24)];

}

//GNU c 的无符号long long 型

int pop64(unsigned int x) {

unsigned long long y;

y = x * 0x0002000400080010ULL;

y = y & 0x1111111111111111ULL;

y = y * 0x1111111111111111ULL;

y = y >> 60;

return y;

}

// 15 位量算术版本

const unsigned __int64 a = 0x0002000400080010;

const unsigned __int64 b = 0x1111111111111111;

int pop15(short x) {

unsigned __int64 y;

y = x * a;

y = y & b;

y = y * b;

y = y >> 60;

return y;

}

// 生成一个包含4 个8 位部分和的字。然后，尽可能把这些字加起来后，生成一个全字和。

// 相加不会产生溢出的8 位字段的字的数目是floor(255/8) = 31

int pop_array(unsigned int A[], int n) {

int i, j, lim;

unsigned int s, s8, x;

s = 0;

for (i = 0; i < n; i += 31) {

lim = __min(n, i + 31);

s8 = 0;

for (j = i; j < lim; j++) {

x = A[j];

x = x - ((x >> 1) & 0x55555555);

x = (x & 0x33333333) + ((x >> 2) & 0x33333333);

x = (x + (x >> 4)) & 0x0F0F0F0F;

s8 = s8 + x;

}

x = (s8 & 0x00FF00FF) + ((s8 >> 8) & 0x00FF00FF);

x = (x & 0x0000FFFF) + (x >> 16);

s = s + x;

}

return s;

}

void test5();

void main()

{

test5();

}

void test5()

{

srand((unsigned)time(NULL));

unsigned int a[256] = {0}; int i;

for (i = 0; i < 256; i++) {a[i] = i;}