1081 Rational Sum （20 分）

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

解析

PAT & MATH：演算法筆記上對分數的處理。

#include<cstdio>
#include<cstdlib>
using namespace std;
using INT64 =
 
 long long;
struct Fraction {
	INT64 up, down;
};
INT64 gcd(INT64 a, INT64 b) {
	if (b == 0)
		return a;
	else
		return gcd(b, a%b);
}
void reproduction(Fraction& a) {
	if (a.down < 0) {
		a.up *= -1;
		a.down *= -1;
	}
	if (a.up == 0)
		a.down = 1;
	else {
		INT64  x = gcd(abs(a.down), abs(a.up));
		a.up /= x;
		a.down /= x;
	}
}
Fraction sum(const Fraction& a, const Fraction& b) {
	Fraction result;
	result.up = a.up*b.down + a.down*b.up;
	result.down = a.down*b.down;
	reproduction(result);
	return result;
}
int main()
{
	int N;
	scanf("%d", &N);
	Fraction base, input;
	scanf("%lld/%lld", &base.up, &base.down);
	for (int i = 1; i < N; i++) {
		scanf("%lld/%lld", &input.up, &input.down);
		base = sum(base, input);
	}
	if (base.up%base.down == 0) {
		printf("%lld", base.up / base.down);
	}
	else {
		if (abs(base.up) > base.down)
			printf("%lld %lld/%lld", base.up / base.down, abs(base.up) % base.down, base.down);
		else
			printf("%lld/%lld", base.up, base.down);
	}
}