题解 | #浮点数加法#

//土尔逊Torson 编写于2023/4/20
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

#define _for(i, a, b) for (int i = a; i < b; ++i)
#define mem(a, b) memset(a, b, sizeof(a))


//有两种方式实现这个精度计算，一种是利用逆置存储计算过程，一种是按照手算计算思维实现

//区别在于 bign change()、bign add()、void print()函数的具体实现方式的不同

/* 第一种实现方式

//将字符串转换为大整数
bign change(string s)
{
    bign c;
    int len = s.length();
    int pos = s.find('.');
    _for(i, 0, len)
    {
        if (i == pos)
            continue;
        i < pos ? c.in[c.ilen++] = s[pos - i - 1] - '0' : c.fn[c.flen++] = s[len - i + pos] - '0';
    }
    return c;
}
//高精度a + b
bign add(bign a, bign b)
{
    bign c;
    int carry = 0;
    //对小数部分进行处理
    _for(i, 0, a.flen)
    {
        int tmp = a.fn[i] + b.fn[i] + carry;
        c.fn[c.flen++] = tmp % 10;
        carry = tmp / 10;
    }
    //对整数部分进行处理
    _for(i, 0, max(a.ilen, b.ilen))
    {
        int tmp = a.in[i] + b.in[i] + carry;
        c.in[c.ilen++] = tmp % 10;
        carry = tmp / 10;
    }
    if (carry != 0)
    {
        c.in[c.ilen++] = carry;
    }
    return c;
}
//输出bign
void print(bign a)
{
    //打印整数部分
    for (int i = a.ilen - 1; i >= 0; --i)
    {
        printf("%d", a.in[i]);
    }
    //打印小数点
    printf(".");
    //打印小数部分
    for (int i = a.flen - 1; i >= 0; --i)
    {
        printf("%d", a.fn[i]);
    }
    printf("\n");
}

*/

// 第二种实现方式
struct bign
{
	int ilen;     //整数部分的长度
	int flen;     //小数部分的长度
	int in[1010]; //整数部分数字
	int fn[1010]; //小数部分数字
	bign()
	{
		ilen = 0, flen = 0;
		mem(in, 0), mem(fn, 0);
	}
};

//让2个浮点数的小数部分长度相同
void cmp(string &a, string &b)
{
	int lena = a.length(), lenb = b.length();
	int posa = a.find('.'), posb = b.find('.');
	lena -= posa, lenb -= posb;
	lena < lenb ? a.append(lenb - lena, '0') : b.append(lena - lenb, '0');
}

//将字符串转换为大整数
bign change(string s)
{
	bign c;
	int len = s.length();
	int pos = s.find('.');
	_for(i, 0, len)
	{
		if (i == pos)
		{
			continue;
		}
		else {
			i < pos ? c.in[c.ilen++] = s[i] - '0' : c.fn[c.flen++] = s[i] - '0';
		}

	}
	return c;
}

//高精度a + b
bign add(bign a, bign b)  // 按照手算带小数的传统计算方式实现精度加法计算
{
	bign c;
	c.flen = max(a.flen, b.flen);
	c.ilen = max(a.ilen, b.ilen);
	int carry = 0;
	//对小数部分进行处理
	int r = (c.flen - 1);

	for(int i = max(a.flen,b.flen) - 1; i >= 0; --i)
	{
		int tmp = a.fn[i] + b.fn[i] + carry;
		c.fn[r--] = tmp % 10;
		carry = tmp / 10;
	}

	//对整数部分进行处理
	int q;
	a.ilen > b.ilen ? q = b.ilen - 1 : q = a.ilen - 1;
	for (int i = max(a.ilen, b.ilen) - 1; i >= 0; --i) {
		if (a.ilen > b.ilen) {
			if (q >= 0) {
				int tmp = a.in[i] + b.in[q--] + carry;
				c.in[i] = tmp % 10;
				carry = tmp / 10;
			}
			else if(q<0){
				int tmp = a.in[i] + carry;
				c.in[i] = tmp % 10;
				carry = tmp / 10;
			}
		}
		else if (a.ilen < b.ilen) {
			if (q >= 0) {
				int tmp = a.in[q--] + b.in[i] + carry;
				c.in[i] = tmp % 10;
				carry = tmp / 10;
			}
			else if(q<0){
				int tmp = b.in[i] + carry;
				c.in[i] = tmp % 10;
				carry = tmp / 10;
			}
		}
		else if (a.ilen == b.ilen) {
			int tmp = a.in[i] + b.in[i] + carry;
			c.in[i] = tmp % 10;
			carry = tmp / 10;
		}
	}

	if (carry != 0)
	{
		c.ilen++;
		int g = c.ilen - 1, h = c.ilen - 2;
		for (int i = c.ilen - 1; i >= 1; i--) {
			c.in[i] = c.in[h--];
		}
		c.in[0] = carry;
	}

	//返回值
	return c;
}

//输出bign
void print(bign a)
{
	//打印整数部分
	for (int i = 0; i <= a.ilen - 1; ++i)
	{
		printf("%d", a.in[i]);
	}
	//打印小数点
	printf(".");
	//打印小数部分
	for (int i = 0; i <= a.flen - 1; ++i)
	{
		printf("%d", a.fn[i]);
	}
	printf("\n");
}

//练习两种方式有助于发散思维
int main()
{
	string s1, s2;
	while (cin >> s1 >> s2)
	{
		cmp(s1, s2);
		bign a = change(s1), b = change(s2);
		print(add(a, b));
	}
	system("pause");
	return EXIT_SUCCESS;
}
// 64 位输出请用 printf("%lld")