1088. Rational Arithmetic (20)

最新推荐文章于 2022-11-01 17:59:59 发布

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#gcd #欧几里德算法 #数字转为string

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本文介绍了一种利用欧几里德算法简化分数并进行分数加减乘除运算的方法，通过字符串到数字的转换处理复杂输入，并提供完整的代码实现。

1.注意在数字和string转化过程中，需要考虑数字不是只有一位的，如300转为“300”，一开始卡在里这里，

测试用例：

24/8 100/10
24/11 300/11

2.该题用到了欧几里德算法求最小公约数gcd(a,b)

算法如下：

//欧几里德算法求最大公约数gcd，其中a>b
long long gcd(long long a, long long b)
{
	return b == 0 ? a : gcd(b, a%b);
}

AC代码如下：

//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;

/*
24/8 100/10
24/11 300/11
*/


//欧几里德算法求最大公约数gcd
long long gcd(long long a, long long b)
{
	return b == 0 ? a : gcd(b, a%b);
}
void str2num(string str, long long&top, long long&bot, int&sign)
{
	bool first = true;
	for (int i = 0; i < str.size(); i++)
	{
		if (str[i] == '-')
			sign = -1;
		else if (str[i] != '/'&&first)
			top = top * 10 + str[i] - '0';
		else if (str[i] == '/')
			first = false;
		else if (str[i] != '/'&&!first)
			bot = bot * 10 + str[i] - '0';
	}
}
string i2s(long long a)
{
	string ans = "";
	if (a == 0) return "0";
	else
	{
		while (a != 0)
		{
			char c = a % 10 + '0';
			ans = c + ans;
			a /= 10;
		}
	}
	return ans;
}
string int2Str(long long top, long long bot, bool sign)
{
	long long tmpGCD = gcd(top, bot);
	top /= tmpGCD;
	bot /= tmpGCD;
	long long tmpInt = top / bot;
	long long tmpRat = top%bot;
	string ans = "";
	if (tmpInt != 0)
	{
		ans = i2s(tmpInt);
	}
	if (tmpRat == 0 && ans.size() != 0)
		;
	else if (tmpRat == 0 && ans.size() == 0)
	{
		return "0";
	}
	else if (tmpRat != 0 && ans.size() != 0)
	{//整数和分数同时存在
		ans += " "+i2s(tmpRat) + "/" + i2s(bot);
	}
	else if (tmpRat != 0 && ans.size() == 0)
	{//仅存在分数
		ans += i2s(tmpRat) + "/" + i2s(bot);
	}
	if (!sign)
		ans = "(-" + ans + ")";
	return ans;
}
int main(void)
{
	string a, b;
	cin >> a >> b;
	long long aTop = 0, aBot = 0, bTop = 0, bBot = 0;
	int aSign = 1;
	int bSign = 1;
	str2num(a, aTop, aBot, aSign);
	str2num(b, bTop, bBot, bSign);
	string aAns;
	string bAns;
	if (aTop != 0)
	{
		int aGCD = gcd(aTop, aBot);
		aTop /= aGCD;
		aBot /= aGCD;
		aAns = int2Str(aTop, aBot, aSign == 1);
	}
	else aAns = "0";
	if (aTop != 0)
	{
		int bGCD = gcd(bTop, bBot);
		bTop /= bGCD;
		bBot /= bGCD;
		bAns = int2Str(bTop, bBot, bSign == 1);
	}
	else 
		bAns = "0";

	//加法：
	long long addBot = aBot*bBot;
	long long addTop = aSign*aTop*bBot + bSign*bTop*aBot;
	bool addSign = (addTop >= 0 ? true : false);
	addTop = labs(addTop);
	long long addInt = addTop / addBot;
	string addAns = int2Str(addTop, addBot, addSign);
	cout << aAns << " + " << bAns << " = " << addAns << endl;

	//减法：
	long long diffBot = aBot*bBot;
	long long diffTop = aSign*aTop*bBot - bSign*bTop*aBot;
	bool diffSign = (diffTop >= 0 ? true : false);
	diffTop = labs(diffTop);
	long long diffInt = diffTop / diffBot;
	string diffAns = int2Str(diffTop, diffBot, diffSign);
	cout << aAns << " - " << bAns << " = " << diffAns << endl;

	//乘法
	long long proBot = aBot*bBot;
	long long proTop = aSign*bSign*aTop*bTop;
	bool proSign = (proTop >= 0 ? true : false);
	proTop = labs(proTop);
	long long proInt = proTop / proBot;
	string proAns = int2Str(proTop, proBot, proSign);
	cout << aAns << " * " << bAns << " = " << proAns << endl;


	//除法
	long long quoBot = aBot*bTop;
	long long quoTop = aSign*bSign*aTop*bBot;
	string quoAns;
	if (quoBot != 0)
	{
		bool quoSign = (quoTop >= 0 ? true : false);
		quoTop = labs(quoTop);
		long long quoInt = quoTop / quoBot;
		quoAns = int2Str(quoTop, quoBot, quoSign);
	}
	else
		quoAns = "Inf";
	cout << aAns << " / " << bAns << " = " << quoAns << endl;

	return 0;
}