高精度运算—利用数组模拟

大整数进行高精度运算，一个基础的方法是开数组进行运算模拟。

高精度加法运算：

#include<iostream>
#include<string>
using namespace std;
string s1, s2;
longlong a[100000], b[100000], c[100000];
int main()
{
	cin >> s1 >> s2;
	

long long lena = s1.size(), lenb = s2.size();

for (long long i = 0; i < lena; i++) //因计算是从右向左进行的，故将数倒序输入到数组中
a[lena - i] = s1[i]-'0';for (long long i = 0; i < lenb; i++)
b[lenb - i] = s2[i]-'0';int lenc = 1, x=0;while (lenc <= lena || lenc <= lenb){c[lenc] = a[lenc] + b[lenc] + x; //加法核心进行进位操作x = c[lenc] / 10;c[lenc] %= 10;lenc++;}c[lenc] = x;if (c[lenc] == 0) //检查最高位lenc--;for (long long i = lenc; i >= 1; i--) cout << c[i];return 0;}

高精度减法运算：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string s1, s2;
long long a[100000], b[100000], c[100000], flag = 0;
int main()
{
	cin >> s1 >> s2;
	long long lena = s1.size(), lenb = s2.size();
	if (lena < lenb || (lena == lenb&&s1 < s2))      //减法需要考虑到减数被减数的大小关系
	{
		swap(lena, lenb);
		string tmp;
		tmp = s1, s1 = s2, s2 = tmp;
		flag = 1;
	}
	for (long long i = 0; i < lena; i++)
		a[lena - i] = s1[i]-'0';
	for (long long i = 0; i < lenb; i++)
		b[lenb - i] = s2[i]-'0';
	int lenc = 1;
	while (lenc <= lena || lenc <= lenb)
	{
		if (a[lenc] < b[lenc])                   //减法核心是进行借位操作
		{ 
			a[lenc + 1]--;
			a[lenc] += 10;
		}
		c[lenc] = a[lenc] - b[lenc];
		lenc++;
	}
	lenc--;
	if (c[lenc] == 0)                                //最高位检测
		lenc--;
	if (s1 != s2)
	{
		if (flag)
			cout << '-';
		for (long long i = lenc; i >= 1; i--)
			cout << c[i];
	}
	else
		cout << '0' << endl;
	system("pause");
	return 0;
}

高精度乘法运算：

#include<iostream>
#include<string>
using namespace std;
string s1, s2;
long long a[100000], b[100000], c[100000];
int main()
{
	cin >> s1 >> s2;
	long long lena = s1.size(), lenb = s2.size();
	for (long long i = 0; i < lena; i++)
		a[lena - i] = s1[i]-'0';
	for (long long i = 0; i < lenb; i++)
		b[lenb - i] = s2[i]-'0';
	for (long long i = 1; i <= lena; i++)
	{
		for (long long j = 1; j <= lenb; j++)        //乘法核心逐位循环相乘
			c[i + j - 1] += a[i] * b[j];
	}
	for (long long i = 1; i <= lena + lenb-1; i++)
	{
		if (c[i] >= 10)                              //进位操作
		{
			c[i + 1] += c[i] / 10;
			c[i] %= 10;
		}
	}
	if(c[lena+lenb]==0)                                  //最高位检测
	    for (long long i =lena+lenb-1; i >= 1; i--)
			cout << c[i];
	else
		for (long long i = lena + lenb; i >= 1; i--)
			cout << c[i];
	return 0;
}