PAT A1088:Rational Arithmetic

题目描述

1088 Rational Arithmetic (20分)
For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and ient.

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 ient 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

求解思路
  • 实现分数的加减乘除,结构体中保存分数的分子和分母,最终打印的时候要分假分数和真分数以不同的形式打印
代码实现(AC)
#includecstdio
#includecmath
#includealgorithm
using namespace std;
typedef long long LL;
struct Fraction{
	LL up;
	LL down;
}f1,f2; 
LL gcd(LL a,LL b)
{
	return !b?a:gcd(b,a%b);
}
Fraction reduction(Fraction result)
{
	if(result.down0)
	{
		result.up=-result.up;
		result.down=-result.down;
	}
	if(result.up==0)	result.down=1;
	else
	{
		int d=gcd(abs(result.up),abs(result.down));
		result.up/=d;
		result.down/=d;
	}
	return result;
}
Fraction add(Fraction f1,Fraction f2)
{
	Fraction result;
	result.up=f1.up*f2.down+f1.down*f2.up;
	result.down=f1.down*f2.down;
	return reduction(result);
}
Fraction minu(Fraction f1,Fraction f2)
{
	Fraction result;
	result.up=f1.up*f2.down-f2.up*f1.down;
	result.down=f1.down*f2.down;
	return reduction(result);
}
Fraction multi(Fraction f1,Fraction f2)
{
	Fraction result;
	result.up=f1.up*f2.up;
	result.down=f1.down*f2.down;
	return reduction(result);
}
Fraction divide(Fraction f1,Fraction f2)
{
	Fraction result;
	result.up=f1.up*f2.down;
	result.down=f1.down*f2.up;
	return reduction(result);
}
void showResult(Fraction r)
{
	r=reduction(r);
	if(r.up0)	printf("(");
	if(r.down==1)	printf("%lld",r.up);
	else if(abs(r.up)r.down)	printf("%lld %lld/%lld",r.up/r.down,(LL)abs(r.up)%r.down,r.down);
	else	printf("%lld/%lld",r.up,r.down);
	if(r.up0)	printf(")");
}
void solve()
{
	scanf("%lld/%lld %lld/%lld",f1.up,f1.down,f2.up,f2.down);
	showResult(f1);printf(" + ");showResult(f2);printf(" = ");showResult(add(f1,f2));printf("\n");
	showResult(f1);printf(" - ");showResult(f2);printf(" = ");showResult(minu(f1,f2));printf("\n");
	showResult(f1);printf(" * ");showResult(f2);printf(" = ");showResult(multi(f1,f2));printf("\n");
	showResult(f1);printf(" / ");showResult(f2);printf(" = ");
	if(f2.up==0)	printf("Inf");
	else showResult(divide(f1,f2));	 
}
int main()
{
	solve();
	return 0;
}
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