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==0) return a;
    else return gcd(b,a%b);    
}
int main()
{
    int N;
    cin >> N;
    long long Inter = 0, resa = 0, resb = 1, a, b, Div, Mul;
    for (int i = 0; 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 == 0 && resa == 0)
        cout << 0 << endl;
    else if (Inter != 0 && resa == 0)
        cout << Inter << endl;
    else if (Inter == 0 && resa != 0)
        cout << resa << "/" << resb << endl;
    else
        cout << Inter << " " << resa << "/" << resb << endl;
    return 0;
}