1088 Rational Arithmetic (20 point(s))

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本文介绍了一种模拟分数加减乘除的算法实现，通过详细解析输入输出规格，展示了如何处理分数运算并保持最简形式。文章提供了完整的C语言代码示例，包括分数的化简、格式处理以及特殊情况的处理，如负数和除数为零的情况。

1088 Rational Arithmetic (20 point(s))

For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient.

Input Specification:

Each input file contains one test case, which gives in one line the two rational numbers in the format a1/b1 a2/b2. The numerators and the denominators are all in the range of long int. If there is a negative sign, it must appear only in front of the numerator. The denominators are guaranteed to be non-zero numbers.

Output Specification:

For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers, respectively. The format of each line is number1 operator number2 = result. Notice that all the rational numbers must be in their simplest form k a/b, where k is the integer part, and a/b is the simplest fraction part. If the number is negative, it must be included in a pair of parentheses. If the denominator in the division is zero, output Inf as the result. It is guaranteed that all the output integers are in the range of long int.

Sample Input 1:

2/3 -4/2

Sample Output 1:

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

Sample Input 2:

5/3 0/6

Sample Output 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

较复杂模拟，分数四则运算。

这道题看似复杂，关键是要能够把控它的共同点，尽可能做到代码复用，避免陷入复杂和混乱。

这里的思路是：四则运算的过程都按照最简单的方法，不考虑任何的化简。只对最后需要输出的结果进行化简和输出格式处理。

#include<stdio.h> 
#define ll long long
ll gcd(ll a,ll b){
	if(b==0) return a;
	else return gcd(b,a%b);
} 
void output(ll a,ll b){//a 分子 b分母 
	if(a==0){
		printf("0");return;
	} 
	bool isNeg = false;
	if(a>0&&b<0){
		isNeg = true; b = -b;
	}else if(a<0&&b>0){
		isNeg = true; a = -a;
	}else if(a<0&&b<0){
		a=-a;b=-b;
	}
	ll f = gcd(a,b);
	a = a/f; b = b/f;
	ll i = a / b;
	a = a - i * b;
	if(isNeg){
		printf("(-");
		if(i!=0){
			printf("%lld",i);
		}
		if(a!=0){
			if(i!=0) printf(" ");
			printf("%lld/%lld",a,b);
		}
		printf(")");
	}else{
		if(i!=0){
			printf("%lld",i);
		}
		if(a!=0){
			if(i!=0) printf(" ");
			printf("%lld/%lld",a,b);
		}		
	}
}
void sum(ll a1,ll b1,ll a2,ll b2){
	ll b = b1*b2;
	ll a = a1*b2+a2*b1;
	output(a1,b1);printf(" + ");
	output(a2,b2);printf(" = ");
	output(a,b);printf("\n");
}
void difference(ll a1,ll b1,ll a2,ll b2){
	ll b = b1*b2;
	ll a = a1*b2-a2*b1;
	output(a1,b1);printf(" - ");
	output(a2,b2);printf(" = ");
	output(a,b);printf("\n");
}
void product(ll a1,ll b1,ll a2,ll b2){
	ll b = b1*b2;
	ll a = a1*a2;
	output(a1,b1);printf(" * ");
	output(a2,b2);printf(" = ");
	output(a,b);printf("\n");
}
void quotient(ll a1,ll b1,ll a2,ll b2){
	output(a1,b1);printf(" / ");
	output(a2,b2);printf(" = ");
	if(a2==0){
		printf("Inf\n");
	}
	else{
		ll b = b1*a2;
		ll a = a1*b2;
		output(a,b);printf("\n");
	}
} 
int main(void){
	ll a1,a2,b1,b2;
	scanf("%lld/%lld",&a1,&b1);
	scanf("%lld/%lld",&a2,&b2);
	sum(a1,b1,a2,b2);
	difference(a1,b1,a2,b2);
	product(a1,b1,a2,b2);
	quotient(a1,b1,a2,b2);
	return 0;
}