PAT甲级——A1081 Rational Sum

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:

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

 #include <iostream>
 #include <vector>
 #include <math.h>
 using namespace std;
 long int  gcd(long int a, long int b)
 {
     if(b==) return a;
     else return gcd(b,a%b);
 }
 int main()
 {
     int N;
     cin >> N;
     long long Inter = , resa = , resb = , a, b, Div, Mul;
     for (int i = ; i < N; ++i)
     {
         char c;
         cin >> a >> c >> b;
         resa = resa * b + a * resb;//同分相加的分子
         resb = resb * b;
         Inter += resa / resb;//简化
         resa = resa - resb * (resa / resb);
         Div = gcd(resb, resa);
         resa /= Div;
         resb /= Div;
     }
     if (Inter ==  && resa == )
         cout <<  << endl;
     else if (Inter !=  && resa == )
         cout << Inter << endl;
     else if (Inter ==  && resa != )
         cout << resa << "/" << resb << endl;
     else
         cout << Inter << " " << resa << "/" << resb << endl;
     return ;
 }