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 (≤), 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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -1/6 1/8

Sample Output 3:

7/24

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 using namespace std; 5 long int  gcd(long int a, long int b) 6 { 7     if(b==0) return a; 8     else return gcd(b,a%b);     9 }10 int main()11 {12     int N;13     cin >> N;14     long long Inter = 0, resa = 0, resb = 1, a, b, Div, Mul;15     for (int i = 0; i < N; ++i)16     {17         char c;18         cin >> a >> c >> b;19         resa = resa * b + a * resb;//同分相加的分子20         resb = resb * b;21         Inter += resa / resb;//简化22         resa = resa - resb * (resa / resb);23         Div = gcd(resb, resa);24         resa /= Div;25         resb /= Div;26     }27     if (Inter == 0 && resa == 0)28         cout << 0 << endl;29     else if (Inter != 0 && resa == 0)30         cout << Inter << endl;31     else if (Inter == 0 && resa != 0)32         cout << resa << "/" << resb << endl;33     else34         cout << Inter << " " << resa << "/" << resb << endl;35     return 0;36 }