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：

在这里插入图片描述

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：
在这里插入图片描述
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：
在这里插入图片描述

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