高精度运算——大数加法与乘法

要点:

• 加法直接传递进位，乘法先保留进位，后统一处理
• 使用int数组存储，空间浪费，处理方便
• 建立bigNum结构（或类），处理清晰方便

代码：

基础定义

#include<bits/stdc++.h>
using namespace std;
char num1[10000];
char num2[10000];
struct bigNum
{
	int num[1000] = {};
	int len;
};
void scan(char* a,bigNum* b)
{
	int l = strlen(a);
	b->len = l;
	for (int i = 0; i < l; i++)
	{
		b->num[i] = a[l - i - 1]-'0';
		//cout << b->num[i];
	}
}

加法

bigNum bigAdd(bigNum* a, bigNum* b)
{
	int carry = 0; int i = 0; int temp = 0;
	int l1 = a->len; int l2 = b->len;
	bigNum res;
	while (i < l1 || i < l2)
	{
		temp = a->num[i] + b->num[i] + carry;
		//cout << temp << endl;
		res.num[i] = temp % 10;
		carry = temp / 10;
		//cout << "c:" << carry << endl;
		i++;
	}
	res.num[i] = carry;
	if (res.num[i] == 0)
		res.len = i+1;
	else
		res.len = i ;
	return res;
}

乘法

bigNum bigMul(bigNum* a, bigNum* b)
{
	bigNum res;
	for (int i = 0; i < a->len; i++)
		for (int j = 0; j < b->len; j++)
			res.num[i + j] += a->num[i] * b->num[j];
	int i = 0;
	while (res.num[i] != 0)
	{
		if (res.num[i] > 9)
		{
			res.num[i + 1] += res.num[i] / 10;
			res.num[i] = res.num[i] % 10;
		}
		i++;
	}
	res.len = i;
	return res;
}

测试主函数

int main()
{
	bigNum a, b;
	cin >> num1 >> num2;
	scan(num1, &a);
	scan(num2, &b);
	bigNum c=bigMul(&a, &b);
	if (c.len == 0)
		cout << 0;
	else
	{
	for (int i = c.len-1; i >= 0; i--)
		cout << c.num[i];
	}
	return 0;
}