详解C/C++高精度算法的简单实现
作者:CodeOfCC
前言
由于上一章《C/C++ 高精度(加减乘除)算法实现》是基于工程项目考虑实现的,也做了一定的优化,实现过程较为复杂。不利于移植和使用,且比较难以理解,时间一长代码也容易忘记,所以重新编写了一个简化的版本,方便以后需要时拷贝使用。
一、基本原理
1、存储方式
采用数字记录高精度数字,数组的第一个元素存储数据长度,比如记录数字为1024示例如下:
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1024+9000:
这里只给出加法的计算说明,其他的以此类推,减法与加法基本一致。乘法和除法略有不同,通过示例图表示也复杂,还不如通过代码去理解,本质的方法就是模拟笔算的立竖式计算。
二、辅助方法
1、字符串转高精度
长度记录在数组第一个元素中
/// <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) { //记录长度 a[0] = strlen(value); for (int i = 1; i <= a[0]; i++) a[i] = value[a[0] - i] - '0'; }
2、整型转高精度
/// <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) { for (size_t i = 1; i < 8096; i++) { a[i] = value % 10; value /= 10; if (!value) { //记录长度 a[0] = i; return; } } }
3、比较
/// <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; }
4、打印
/// <summary> /// 打印输出结果 /// </summary> static void print(int* a) { if (!a[0]) printf("0"); for (int i = a[0]; i > 0; i--) printf("%d", a[i]); }
三、算法实现
原理就不做具体介绍了,四种计算的核心都是模拟立竖式计算。
1、加法
为了保证代码相对简单,当b长度较小时可能会做一些多余的计算,不影响结果。
/// <summary> /// 加法(累加) ///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 acc(int* a, int* b) { int len = a[0] > b[0] ? a[0] : b[0]; memset(a + a[0] + 1, 0, (len - a[0] + 1) * sizeof(int)); memset(b + b[0] + 1, 0, (len - b[0] + 1) * sizeof(int)); for (int i = 1; i <= len; i++) { int temp = a[i] + b[i]; a[i] = temp % 10; a[i + 1] += temp / 10; } if (a[len + 1])a[0]++; }
2、减法
/// <summary> /// 减法(累减) ///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 subc(int* a, int* b) { memset(b + b[0] + 1, 0, (a[0] - b[0]) * sizeof(int)); for (int i = 1; i <= a[0]; i++) { int temp = a[i] - b[i]; a[i] = temp; if (temp < 0) { //借位 a[i + 1] -= 1; a[i] += 10; } } //记录长度 for (int i = a[0]; i > 0; i--) if (a[i]) { a[0] = i; return; } a[0] = 0; }
3、乘法
/// <summary> /// 乘法 /// </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[]) { c[a[0] + b[0]] = 0; memset(c, 0, sizeof(int) * (a[0] + b[0] + 1)); for (int i = 1; i <= a[0]; i++) { int j; int d = 0; //被乘数的一位去乘以乘数的每一位 for (j = 1; j <= b[0]; j++) { int temp = a[i] * b[j] + c[j + i - 1] + d; c[j + i - 1] = temp % 10; d = temp / 10; } if (d) { c[j + i - 1] = d; } } //记录长度 for (int i = a[0] + b[0]; i > 0; i--) if (c[i]) { c[0] = i; return; } }
4、除法
采用了升阶+减法实现
/// <summary> /// 除法 /// 依赖减法subc /// </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aLen-bLen+1</param> /// <param name="mod">[out]余数,数组长度大于等于aLen</param>> /// <param name="temp">[in]临时缓冲区,由外部提供以提高性能,数组长度大于等于aLen-bLen+1</param> static void divi(int* a, int* b, int* c, int* mod, int* temp) { //相差的阶数 int digit = a[0] - b[0] + 1; memcpy(mod, a, (a[0] + 1) * sizeof(int)); memset(c, 0, sizeof(int) * (digit + 1)); memset(temp, 0, sizeof(int) * digit); while (digit) { //升阶 memcpy(temp + digit, b + 1, sizeof(int) * b[0]); temp[0] = b[0] + digit - 1; //减法 while (compare(mod, temp) != -1) { subc(mod, temp); c[digit]++; } digit--; } //记录长度 for (int i = a[0] - b[0] + 1; i > 0; i--) if (c[i]) { c[0] = i; return; } }
四、使用示例
1、加法
计算累加
int main() { int64_t n; int num[1024]; int num2[1024]; std::cin >> n; loadInt(num, 0); for (int64_t i = 1; i <= n; i++) { loadInt(num2, i); acc(num, num2); } print(num); return 0; }
结果:
2、减法
两个任意n位数的减法,数字1大于数字2。
int main() { int a1[8096], a2[8096]; std::string s1, s2; std::cin >> s1 >> s2; loadStr(a1, s1.c_str()); loadStr(a2, s2.c_str()); subc(a1, a2); print(a1); return 0; }
结果:
#数字1
752425289999999999999652142141414141414146666676667677682324000001302461646520
#数字2
587891851201874512000000000154515100202121555555555555555555555545477910232111
#计算结果
164533438798125487999652141986899041212025111121112122126768444455824551414409
3、乘法
计算阶乘
int main() { int64_t n; int num[8192]; int num2[8192]; int num3[8192]; int* p1 = num; int* p2 = num3; std::cin >> n; loadInt(num, 1); for (int64_t i = 1; i <= n; i++) { loadInt(num2, i); mul(p1, num2, p2); int* temp = p1; p1 = p2; p2 = temp; } print(p1); return 0; }
结果:
#阶乘数
1000
#计算结果
402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
4、除法
给定两个非负整数A,B,请你计算 A / B的商和余数。
int main() { int a1[8096], a2[8096], c[8096], mod[8096], temp[8096]; std::string s1, s2; std::cin >> s1 >> s2; loadStr(a1, s1.c_str()); loadStr(a2, s2.c_str()); divi(a1, a2, c, mod, temp); print(c); std::cout << std::endl; print(mod); return 0; }
结果:
#被除数
12458848948151231366666666666666665454545123156415641561231561213648
#除数
88484851521548496564154848456486789
#商
140802055198308817458997123299946
#余数
25178368711335236611547594127800254
总结
以上就是今天要讲的内容,本文提供的是较为简化的实现,且每个方法基本是独立的,可单独拿来使用,用法也比较简单,由于采用数组第一个元素存储长度,接口就变得很简洁,使用起来也方便了很多。
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