1034. 有理数四则运算(20)

本题要求编写程序，计算2个有理数的和、差、积、商。

输入格式：

输入在一行中按照“a1/b1 a2/b2”的格式给出两个分数形式的有理数，其中分子和分母全是整型范围内的整数，负号只可能出现在分子前，分母不为0。

输出格式：

分别在4行中按照“有理数1 运算符 有理数2 = 结果”的格式顺序输出2个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式“k a/b”，其中k是整数部分，a/b是最简分数部分；若为负数，则须加括号；若除法分母为0，则输出“Inf”。题目保证正确的输出中没有超过整型范围的整数。
输入样例1：

2/3 -4/2

输出样例1：

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

输入样例2：

5/3 0/6

输出样例2：

1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

</pre><p><pre name="code" class="cpp">#include<iostream>
#include<cmath>
#include<cstring>

using namespace std;
#define LLI long long int
#define MAX 50

/**计算最大公约数**/
LLI computeGCD(LLI num1, LLI num2){
if(num1 < 0){
num1 = num1*(-1);
}
if(num2 < 0){
num2 = num2*(-1);
}
while(num1){
if(num1<num2){
LLI t = num1;
num1 = num2;
num2 = t;
}
num1 = num1%num2;
}
return num2;
}

/**改变分子和分母的正负，负数统一在分子**/
void changeNegative(LLI *a, LLI *b){
if(*a>0 && *b>0){
return;
}else if((*a>0 && *b<0)){
*a = *a*(-1);
*b = abs(*b);
return;
}else if(*a<0 && *b<0){
*a = abs(*a);
*b = abs(*b);
return;
}
}

/**
*转换为真分数输出
*param:dividend:被除数 divisor:除数
**/
void changeToProperFraction(LLI dividend, LLI divisor){
LLI remainder = 0; //余数
LLI dealer = 0; //商
LLI gcd = 1; //最大公约数
if((gcd = computeGCD(dividend, divisor)) != 1){ //判断是否可约
dividend = dividend/gcd;
divisor = divisor/gcd;
}
remainder = dividend%divisor;
dealer = dividend/divisor;
if(dividend == 0){ //情况1:被除数为0 eg:0/3
cout<<dividend;
}else if(dealer == 0){ //情况2:商为0 eg:1/3 2/4 -2/4
if(dividend < 0){
cout<<"("<<dividend<<"/"<<abs(divisor)<<")";
}else{
cout<<dividend<<"/"<<abs(divisor);
}
}else if(remainder == 0){ //情况3:余数为0 eg:4/2
if(dividend < 0){
cout<<"("<<dealer<<")";
}else{
cout<<dealer;
}
}else if(remainder != 0){ //情况4:余数不为0 eg:4/3 8/6 -8/6
if(dividend < 0){
cout<<"(";
if(dealer != 0){
cout<<dealer<<" ";
}
cout<<abs(remainder)<<"/"<<abs(divisor)<<")";
}else{
if(dealer != 0){
cout<<dealer<<" ";
}
cout<<abs(remainder)<<"/"<<abs(divisor);
}
}
}

/**将输入的分数字符串改为对应的数字**/
void changeToNum(char str1[], char str2[], LLI *a1, LLI *b1, LLI *a2, LLI *b2){
int i = 0, j = 0;
while(i<strlen(str1) && str1[i]!='/'){
if(str1[i] == '-'){
i++;
}
*a1 = *a1*10;
*a1 += str1[i++]-'0';
}
i++;
while(i<strlen(str1)){
*b1 = *b1*10;
*b1 += str1[i++]-'0';
}
while(j<strlen(str2) && str2[j]!='/'){
if(str2[j] == '-'){
j++;
}
*a2 = *a2*10;
*a2 += str2[j++]-'0';
}
j++;
while(j<strlen(str2)){
*b2 = *b2*10;
*b2 += str2[j++]-'0';
}
if(str1[0] == '-'){
*a1 = *a1*(-1);
}
if(str2[0] == '-'){
*a2 = *a2*(-1);
}
}

int main(){
char str1[MAX],str2[MAX];
LLI a1 = 0, b1 = 0, a2 = 0, b2 = 0;
LLI a3 = 0, b3 = 0; //用于存放四则运算的结果

cin>>str1>>str2;
changeToNum(str1, str2, &a1, &b1, &a2, &b2); //将输入改为数字

//+
changeToProperFraction(a1,b1);
cout<<" + ";
changeToProperFraction(a2,b2);
a3 = a1*b2+a2*b1;
b3 = b1*b2;
cout<<" = ";
changeNegative(&a3, &b3);
changeToProperFraction(a3,b3);
cout<<endl;
//-
changeToProperFraction(a1,b1);
cout<<" - ";
changeToProperFraction(a2,b2);
a3 = a1*b2-a2*b1;
b3 = b1*b2;
cout<<" = ";
changeNegative(&a3, &b3);
changeToProperFraction(a3,b3);
cout<<endl;
//*
changeToProperFraction(a1,b1);
cout<<" * ";
changeToProperFraction(a2,b2);
a3 = a1*a2;
b3 = b1*b2;
cout<<" = ";
changeNegative(&a3, &b3);
changeToProperFraction(a3,b3);
cout<<endl;

changeToProperFraction(a1,b1);
cout<<" / ";
changeToProperFraction(a2,b2);
a3 = a1*b2;
b3 = b1*a2;
cout<<" = ";
changeNegative(&a3, &b3);
if(b3 == 0){
cout<<"Inf"<<endl;
}else{
changeToProperFraction(a3,b3);
cout<<endl;
}
return 0;
}