笔试题之字符串大数相减和大数相乘

1. 大数相减

首先判断符号，然后再计算大小

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

string Sub(string , string );
string sub(string a, string b)
{
	// 1. judge who is biger
	bool firstBig = false;
	if (a.size() > b.size())
	{
		firstBig = true;
	}
	else if (a.size() == b.size())
	{
		int i = 0;
		while (i < a.size() && a[i] == b[i])
			i++;
		if (i == a.size() || a[i] == b[i]) return string(1,'0');
		if (a[i] > b[i])  firstBig = true;
	}
        // 2. get the output
	string c;
	if (firstBig == true)
		return Sub(a, b);
	else
		return '-' + Sub(b, a);
}

string Sub(string a, string b)
{
	int dec = 0;
	int m = a.size() - 1;
	int n = b.size() - 1;
	string c(a.size(), '0');
	while (m >= 0 && n >= 0)
	{
		int temp = a[m] - b[n] - dec;
		if (temp < 0)
		{
			dec = 1;
			temp += 10;
		}
		c[m] = temp + '0';
		m--;
		n--;
	}
	while (m >= 0)
	{
		int temp = a[m] - dec - '0';
		if (temp < 0)
		{
			dec = 1;
			temp += 10;
		}
		c[m] = temp + '0';
		m--;
	}
	m = 0;
	while (c[m] == '0')
	{
		m++;
	}
	return c.substr(m, c.size() - m);
}

int main()
{
	string a, b;
	while (cin >> a >> b)
	{
		cout << sub(a, b) <<endl;
	}
	return 0;
}

2. 大数相乘

与手算乘法的流程一致，首先计算有多少个1，多少个10，多少个100，最后再重新统计表示

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;

string muti(string a, string b)
{
	if (a.size() == 0 || b.size() == 0 || a == "0" || b == "0") return "0";
	vector<int> ret(a.size() + b.size(),0);
	int base = a.size() + b.size() - 2;
	for (int i = b.size() - 1; i >= 0; i--)
	{
		for (int j = a.size() - 1; j >= 0; j--)
		{
			ret[base - i - j] += (a[j] - '0') * (b[i] - '0');
		}
	}
	int carry = 0;
	for (int i = 0; i < ret.size(); i++)
	{
		int tmpcarry = (ret[i] + carry) / 10;
		ret[i] = (ret[i] + carry) % 10;
		carry = tmpcarry;
	}
	while (carry > 0)
	{
		ret.push_back(carry%10);
		carry = carry / 10;
	}
	
	int i = ret.size() - 1;

	while (ret[i] == 0)
		i--;
	string c(i+1, '0');
	for (int j = 0; i >= 0; i--)
	{
		c[i] = ret[j++] + '0';
	}
	return c;

}

int main()
{
	string a, b;
	while (cin >> a >> b)
	{
		cout << muti(a, b) <<endl;
	}
	return 0;
}