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
题意：求几个分数的和，结果化为最简分数。
解题思路：
1.分数计算主要问题在于求最大公约数和最小公倍数，最大公约数可以使用辗转相除法，最小公倍数可由两数乘积/最大公倍数得到。
2.两个分数相加，先要把两个分母的最小公倍数求出来，也就是通分，相加之后再对和进行化简，也是就求分子分母的最大公约数。
3.多个数相加就是循环调用两个分数相加的函数。
4.最后对结果进行格式化输出，需要几个判断，有分子为0，分母为1，分子大于分母，以及分子为负数的情况。
n = int(input())
l = input().split()
#l = ['2/5', '4/15', '1/30', '-2/60', '8/3']
#l = ['4/3', '2/3']
#l = ['1/3', '-1/6', '1/8']
numerators = []
denominators = []
def gcd(n1,n2):#greatest common divisor
return gcd(n2, n1 % n2) if n2 else n1
# if n2:
# return gcd(n2, n1 % n2)
# else:
# return n1
def lcm(n1,n2):#lowest common multiple
return n1*n2//gcd(n1,n2)
def rational_sum(r1, r2):
n1, d1 = map(int,r1.split('/'))
n2, d2 = map(int,r2.split('/'))
d3 = lcm(d1, d2)
n3 = n1*(d3//d1) + n2*(d3//d2)
if n3:
gcd1 = gcd(abs(n3),d3)
d3 = d3//gcd1
n3 = n3//gcd1
return str(n3)+'/'+str(d3)
s = l[0]
for i in range(1,len(l)):
s = rational_sum(s,l[i])
#print(s)
n1, d1 = map(int,s.split('/'))
if d1 == 1:
print(n1)
elif n1 == 0:
print(n1)
elif abs(n1) > d1:
if n1>0:
print('%d %d/%d'%(n1//d1,n1%d1,d1))
else:
print('%d %d/%d'%((n1//d1)+1,abs(n1)%d1,d1))
else:
print(s)