前言
由于上一章《C/C++ 高精度（加减乘除）算法简单实现》实现了基本的高精度计算，数组的每个元素存储一位10进制的数字。这样的存储方式并不是最优的，32位的整型其实至少可以存储9位高精度数字，数组元素存储更多的位数就是压位优化。本文将展示压位优化的原理以及压9位的实现和性能对比。
一、基本原理
1、存储方式
压位优化就是将原本存储一位的数组元素变成存储多位，这样就可以提升运算效率，通常最高能存储9位，如下图示例为存储4位。
2、计算方式
采用模拟立竖式计算，比如加法的计算流程，如下图所示，20481024+80001000=100482024：
二、完整代码
因为接口以及使用方法与上一章《C/C++ 高精度（加减乘除）算法简单实现》是完全一致的，所以这里直接展示完整代码，省略使用示例。下面代码为压9位实现。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdint.h>
/// <summary>
/// 通过字符串初始化
/// </summary>
/// <param name="a">[in]高精度数组</param>
/// <param name="value">[in]字符串首地址</param>
static void loadStr(int* a, const char* value) {
int len = strlen(value);
int left = len % 9;
char s[8], * p = (char*)value + left;
a[0] = ceil(len / 9.0);
len = len / 9.0;
for (int i = 1; i <= len; i++)
sscanf(p + (len - i ) * 9, "%09d", &a[i]);
if (left){
sprintf(s, "%%0%dd", left);
sscanf(value, s, &a[a[0]]);
}
}
/// <summary>
/// 输出到字符串，
/// </summary>
/// <param name="a">[in]高精度数组</param>
/// <param name="str">[out]字符串，由外部停供缓冲区，需要保证长度足够</param>
static void toStr(int* a, char* str) {
if (!a[0]) {
sprintf(str, "0");
return;
}
sprintf(str, "%d", a[a[0]]);
str += strlen(str);
for (int i = a[0]-1; i > 0; i--)
sprintf(str +( a[0] -i-1)*9, "%09d", a[i]);
str[(a[0]-1)*9] = '\0';
}
/// <summary>
/// 通过无符号整型初始化
/// </summary>
/// <param name="a">[in]高精度数组</param>
/// <param name="value">[in]整型值</param>
static void loadInt(int* a, uint64_t value) {
a[0] = 0;
while (value)a[++a[0]] = value % 1000000000, value /= 1000000000;
}
/// <summary>
/// 比较两个高精度数的大小
/// </summary>
/// <param name="a">[in]第一个数</param>
/// <param name="b">[in]第二个数</param>
/// <returns>1是a>b,0是a==b,-1是a<b</returns>
static int compare(int* a, int* b) {
if (a[0] > b[0])return 1;
if (a[0] < b[0])return -1;
for (int i = a[0]; i > 0; i--)
if (a[i] > b[i])return 1;
else if (a[i] < b[i])return -1;
return 0;
}
/// <summary>
/// 复制
/// </summary>
/// <param name="a">[in]源</param>
/// <param name="b">[in]目标</param>
static void copy(int* a, int* b) {
memcpy(b, a, (a[0] + 1) * sizeof(int));
}
/// <summary>
/// 打印输出结果
/// </summary>
static void print(int* a) {
int i = a[0];
printf("%d", a[i--]);
for (; i > 0; i--)printf("%09d", a[i]);
}
/// <summary>
/// 加
/// </summary>
/// <param name="a">[in]被加数</param>
/// <param name="b">[in]加数</param>
/// <param name="c">[out]结果</param>
static void plus(int* a, int* b, int* c) {
int* p;
if (a[0] < b[0])p = a, a = b, b = p;//确保a长度最大
int i = 1, alen = a[0], blen = b[0];
c[0] = c[alen + 1] = 0;
if (a != c)memcpy(c + blen + 1, a + blen + 1, (alen - blen) * sizeof(int));//a多出的部分直接拷贝到结果
for (; i <= blen; i++) {
c[i] = a[i] + b[i];
if (c[i - 1] >= 1000000000)c[i - 1] -= 1000000000, c[i]++;//判断上一位是否进位
}
i--;
while (c[i++] >= 1000000000)c[i - 1] -= 1000000000, c[i]++;//继续判断进位
c[0] = c[alen + 1] ? alen + 1 : alen;//记录长度
}
/// <summary>
/// 加等于
///结果会保存在a中
/// </summary>
/// <param name="a">[in]被加数</param>
/// <param name="b">[in]加数</param>
static void plusq(int* a, int* b) {
plus(a, b, a);
}
/// <summary>
/// 减
/// </summary>
/// <param name="a">[in]被减数,被减数必须大于等于减数</param>
/// <param name="b">[in]减数</param>
/// <param name="c">[out]结果</param>
static void sub(int* a, int* b, int* c) {
int i = 1, alen = a[0];
if (a != c)memcpy(c + b[0] + 1, a + b[0] + 1, (a[0] - b[0]) * sizeof(int));//a多出的部分直接拷贝到结果
c[0] = 1;
for (; i <= b[0]; i++) {
c[i] = a[i] - b[i];
if (c[i - 1] < 0)c[i - 1] += 1000000000, c[i] --;//判断上一位是否补位
}
i--;
while (c[i++] < 0)c[i - 1] += 1000000000, c[i]--;//继续判断补位
while (!c[alen--]); c[0] = alen + 1;//记录长度
}
/// <summary>
/// 减法等于
///结果会保存在a中
/// </summary>
/// <param name="a">[in]被减数，被减数必须大于等于减数</param>
/// <param name="b">[in]减数</param>
static void subq(int* a, int* b) {
sub(a, b, a);
}
/// <summary>
/// 乘
/// </summary>
/// <param name="a">[in]被乘数</param>
/// <param name="b">[in]乘数</param>
/// <param name="c">[out]结果，数组长度必须大于等于aLen+bLen+1</param>
static void mul(int* a, int* b, int c[]) {
int len = a[0] + b[0], d = 0;
memset(c, 0, sizeof(int) * (len + 1));
b[b[0] + 1] = 0; c[0] = 1;//防止越界
for (int i = 1; i <= a[0]; i++)
for (int j = 1; j <= b[0] + 1; j++){
int64_t t = (int64_t)a[i] * b[j] + c[j + i - 1] + d;
c[j + i - 1] = t % 1000000000;
d = t / 1000000000;
}
while (!c[len])len--; c[0] = len;
}
/// <summary>
/// 乘等于
/// 累乘,结果存放于a
/// </summary>
/// <param name="a">[in]被乘数，数组长度必须大于等于2aLen+bLen+1</param>
/// <param name="b">[in]乘数</param>
static void mulq(int* a, int* b) {
int* c = a + a[0] + b[0] + 1;
memcpy(c, a, (a[0] + 1) * sizeof(int));
mul(c, b, a);
}
/// <summary>
/// 除法
/// 依赖减法subq
/// </summary>
/// <param name="a">[in]被除数,被除数必须大于除数</param>
/// <param name="b">[in]除数</param>
/// <param name="c">[out]商，数组长度大于等于3aLen-bLen+1</param>
/// <param name="mod">[out]余数，可以为NULL，数组长度大于等于aLen</param>>
static void div(int* a, int* b, int* c, int* mod) {
int len = a[0] - b[0] + 1, times, hTimes[32], * temp = c + a[0] + 1;
if (!mod)mod = temp + 2*(a[0] + 1)+1;//缓冲区
memcpy(mod, a, (a[0] + 1) * sizeof(int));
memset(c, 0, sizeof(int) * (len + 1));
memset(temp, 0, sizeof(int) * len);
c[0] = 1;//防止while越界
for (int i = len; i > 0; i--) {
memcpy(temp + i, b + 1, sizeof(int) * b[0]);//升阶
temp[0] = b[0] + i - 1;
while (compare(mod, temp) != -1) {
if (times = (mod[mod[0]] * ((mod[0] - temp[0]) ? 1000000000ll : 1)) / (temp[temp[0]] + (temp[0] == 1 ? 0 : 1)))//升倍数
{
loadInt(hTimes,times);
mulq(temp, hTimes);
}
else times = 1;
while (compare(mod, temp) != -1)subq(mod, temp), c[i] += times; //减法
memcpy(temp + i, b + 1, sizeof(int) * b[0]);//还原
temp[0] = b[0] + i - 1;
}
}
while (!c[len])len--; c[0] = len;
}
/// <summary>
/// 除等于
/// 商保存在a
/// 依赖div
/// </summary>
/// <param name="a">[in]被除数,被除数必须大于除数</param>
/// <param name="b">[in]除数</param>
/// <param name="mod">[out]余数，可以为NULL，数组长度大于等于aLen</param>>
static void divq(int* a, int* b, int* mod) {
div(a, b, a, mod);
}
三、性能对比
测试平台：Windows 11
测试设备：i7 8750h
测试方式：测试5次取均值
表1、测试用例
| 测试用例 | 描述 |
|---|---|
| 1 | 整型范围数字计算500000次 |
| 2 | 长数字与整型范围数字计算500000次 |
| 3 | 长数字与长数字计算500000次 |
基于上述用例编写程序进行测试，测试结果如下表
表2、测试结果
| 计算 | 测试用例 | 1位实现（上一章）耗时 | 9位优化（本章）耗时 |
|---|---|---|---|
| 加法 | 测试用例1 | 0.003926s | 0.002620s |
| 加法 | 测试用例2 | 0.026735s | 0.005711s |
| 加法 | 测试用例3 | 0.029378s | 0.005384s |
| 累加 | 测试用例1 | 0.003255s | 0.002536s |
| 累加 | 测试用例2 | 0.017843s | 0.002592s |
| 累加 | 测试用例3 | 0.034025s | 0.006474s |
| 减法 | 测试用例1 | 0.004237s | 0.002078s |
| 减法 | 测试用例2 | 0.024775s | 0.004939s |
| 减法 | 测试用例3 | 0.027634s | 0.004929s |
| 累减 | 测试用例1 | 0.004272s | 0.002034s |
| 累减 | 测试用例2 | 0.005407 | 0.001942s |
| 累减 | 测试用例3 | 0.019363s | 0.004282s |
| 乘法 | 测试用例1 | 0.043608s | 0.004751s |
| 乘法 | 测试用例2 | 0.479071s | 0.028358s |
| 乘法 | 测试用例3 | 3.375447s | 0.064259s |
| 累乘 | 测试用例1 只计算1000次 | 0.001237s | 0.000137s |
| 累乘 | 测试用例2 只计算1000次 | 0.001577s | 0.000187s |
| 累乘 | 测试用例3 只计算1000次 | 5.792887s | 0.081988s |
| 除法 | 测试用例1 | 0.025391s | 0.024763s |
| 除法 | 测试用例2 | 5.292809s | 0.516090s |
| 除法 | 测试用例3 | 0.395773s | 0.073812s |
| 累除 | 测试用例1 只计算1000次 | 0.059054s | 0.035722s |
| 累除 | 测试用例2 只计算1000次 | 0.103727s | 0.060936s |
| 累除 | 测试用例3 只计算500次 | 89.748837s | 25.126072s |
将上表数据进行分类相同类型取均值计算出提升速度如下图所示，仅作参考。
图1、速度提升
总结
以上就是今天要讲的内容，压位优化性能提升是比较显著的，而且实现也很容易，大部分逻辑是一致的只是底数变大了而已。从性能测试结果来看所有计算至少由4倍的提升，乘法性能提升较大有可能是测试方法不严谨，这个待以后验证。总的来说，对高精度运算进行压位优化还是很有必要的，尤其是对时间和空间有要求的场景还是比较适用的。
附录
1、性能测试代码
#include<Windows.h>
#include <iostream>
static int a[819200];
static int b[819200];
static int c[819200];
static int mod[819200];
static char str[81920];
/// <summary>
/// 返回当前时间
/// </summary>
/// <returns>当前时间，单位秒，精度微秒</returns>
static double getCurrentTime()
{
LARGE_INTEGER ticks, Frequency;
QueryPerformanceFrequency(&Frequency);
QueryPerformanceCounter(&ticks);
return (double)ticks.QuadPart / (double)Frequency.QuadPart;
}
/// <summary>
/// 性能测试
/// </summary>
static void test() {
double d = getCurrentTime();
loadStr(a, "50000");
loadInt(b, 50000);
for (int64_t i = 1; i <= 500000; i++) {
plus(a, b, c);
}
printf("plus performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadInt(b, 5);
for (int64_t i = 1; i <= 500000; i++) {
plus(a, b, c);
}
printf("plus performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 500000; i++) {
plus(b, a, c);
}
printf("plus performence 3: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "50000");
loadInt(b, 50000);
for (int64_t i = 1; i <= 500000; i++) {
plusq(a, b);
}
printf("plusq performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
for (int64_t i = 500000000; i <= 500000000 + 500000; i++) {
loadInt(b, i);
plusq(a, b);
}
printf("plusq performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(b, "999999999999999999999999999999999999999999999999999999999999999999");
for (int64_t i = 500000000; i <= 500000000 + 500000; i++) {
loadInt(a, i);
plusq(a, b);
}
printf("plusq performence 3: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "50000");
loadInt(b, 10000);
for (int64_t i = 1; i <= 500000; i++) {
sub(a, b, c);
}
printf("sub performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "100000000000000000000000000000000000000000000000000000000000000000");
loadInt(b, 11111);
for (int64_t i = 1; i <= 500000; i++) {
sub(a, b, c);
}
printf("sub performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "100000000000000000000000000000000000000000000000000000000000000000");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 500000; i++) {
sub(a, b, c);
}
printf("sub performence 3: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "50000000000");
loadInt(b, 500000);
for (int64_t i = 1; i <= 500000; i++) {
subq(a, b);
}
printf("subq performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "100000000000000000000000000000000000000000000000000000000000000000");
loadInt(b, 11111);
for (int64_t i = 1; i <= 500000; i++) {
subq(a, b);
}
printf("subq performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "100000000000000000000000000000000000000000000000000000000000000000");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 500000; i++) {
subq(a, b);
}
printf("subq performence 3: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "50000");
loadInt(b, 12345);
for (int64_t i = 1; i <= 500000; i++) {
mul(a, b, c);
}
printf("mul performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadInt(b, 12345);
for (int64_t i = 1; i <= 500000; i++) {
mul(a, b, c);
}
printf("mul performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 500000; i++) {
mul(b, a, c);
}
printf("mul performence 3: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "2");
loadInt(b, 2);
for (int64_t i = 1; i <= 1000; i++) {
mulq(a, b);
}
printf("mulq performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadInt(b, 2);
for (int64_t i = 1; i <= 1000; i++) {
mulq(a, b);
}
printf("mulq performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 1000; i++) {
mulq(b, a);
}
printf("mulq performence 3: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "50000");
loadInt(b, 12345);
for (int64_t i = 1; i <= 500000; i++) {
div(a, b, c, mod);
}
printf("div performence 1: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "100000000000000000000000000000000000000000000000000000000000000000");
loadInt(b, 12345);
for (int64_t i = 1; i <= 500000; i++) {
div(a, b, c, NULL);
}
printf("div performence 2: %llfs\n", getCurrentTime() - d);
d = getCurrentTime();
loadStr(a, "100000000000000000000000000000000000000000000000000000000000000000");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 500000; i++) {
div(a, b, c, mod);
}
printf("div performence 3: %llfs\n", getCurrentTime() - d);
loadStr(a, "1");
loadStr(b, "2");
for (int64_t i = 1; i <= 1000; i++) {
mulq(a, b);
}
d = getCurrentTime();
for (int64_t i = 1; i <= 1000; i++) {
divq(a, b, mod);
}
printf("divq performence 1: %llfs\n", getCurrentTime() - d);
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadStr(b, "2");
for (int64_t i = 1; i <= 1000; i++) {
mulq(a, b);
}
d = getCurrentTime();
for (int64_t i = 1; i <= 1000; i++) {
divq(a, b, mod);
}
printf("divq performence 2: %llfs\n", getCurrentTime() - d);
loadStr(a, "999999999999999999999999999999999999999999999999999999999999999999");
loadStr(b, "11111111111111111111111111111111111111");
for (int64_t i = 1; i <= 500; i++) {
mulq(a, b);
}
d = getCurrentTime();
for (int64_t i = 1; i <= 500; i++) {
divq(a, b, mod);
}
printf("divq performence 3: %llfs\n", getCurrentTime() - d);
}