PAT 1081. Rational Sum (20)

http://pat.zju.edu.cn/contests/pat-a-practise/1081

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

题目难度不大，但是这种题目真的是讨厌，各种小问题，要反复调试。

#include <cstdio>
long long gcd(long long x, long long y){
	if (y == 0) return x;
	else return gcd(y, x%y);
}
long long lcm(long long x, long long y){
	return x / gcd(x, y) *y;
}
long long a[100], b[100];
int main(){
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++){
		scanf("%lld/%lld", &a[i], &b[i]);
	}
	long long deno = 1, nume = 0;
	for (int i = 0; i < n; i++)
		deno = lcm(deno, b[i]);
	for (int i = 0; i < n; i++)
		nume += deno / b[i] * a[i];
	long long pre = nume / deno;
	nume = nume - pre*deno;
	long long t = gcd(nume, deno);
	nume = nume / t;
	deno = deno / t;
	if (pre == 0 && nume == 0)
		printf("0\n");
	else if (pre != 0 && nume == 0)
		printf("%lld\n", pre);
	else if (pre == 0 && nume != 0)
		printf("%lld/%lld\n", nume, deno);
	else 
		printf("%lld %lld/%lld\n", pre, nume, deno);
	return 0;
}