PAT (Advanced Level) Practice 1081 Rational Sum (20分) （注意精度问题！）

1.题目

Given N rational numbers in the form numerator/denominator, you are supposed to calculate their sum.

Input Specification:

Each input file contains one test case. Each case starts with a positive integer N (≤100), followed in the next line N rational numbers a1/b1 a2/b2 ... where all the numerators and denominators are in the range of long int. If there is a negative number, then the sign must appear in front of the numerator.

Output Specification:

For each test case, output the sum in the simplest form integer numerator/denominator where integer is the integer part of the sum, numerator < denominator, and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.

Sample Input 1:

5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

2
4/3 2/3

Sample Output 2:

2

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

2.代码

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
long long int a1, a2;
long long int add1=0, add2=1;
int gcd(long long int a, long long int b)
{
	return b == 0 ? a : gcd(b, a%b);
}
void output(long long int tempint, long long int tempa, long long int tempb)
{
	if (tempint == 0)
	{
		if (tempa > 0)printf("%lld/%lld", tempa, tempb);
		else if (tempa == 0)printf("0");
	}
	else
	{
		if (tempa > 0)printf("%lld %lld/%lld", tempint, tempa, tempb);
		else if (tempa == 0)printf("%lld", tempint);
	}


}
int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%lld/%lld", &a1, &a2);
		long long int d = gcd(a1, a2);
		a1 = a1 / d;
		a2 = a2 / d;
		add1 = add1*a2 + add2*a1;
		add2 = add2*a2;
		d = gcd(add1, add2);
		add1 = add1 / d;
		add2 = add2 / d;
	}
	long long int inte = add1 / add2;
	add1 = abs(add1);
	add1 = add1 - inte*add2;
	output(inte, add1, add2);
	printf("\n");
}