详解C/C++高精度(加减乘除)算法中的压位优化
CodeOfCC 人气:0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1000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重,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); }
加载全部内容