1081. Rational Sum (20)

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

#include <iostream>
#include <math.h>
using namespace std;
void jia(long &a,long  &b,long  ax,long  bx);
void out(long a,long  b);
int main()
{
  char s;
  int n,i;
  long  ax,bx,a,b;
  cin>>n;
  cin>>a>>s>>b;
  for(i=1;i<n;i++)
  {
    cin>>ax>>s>>bx;
    jia(a,b,ax,bx);  
  }
  out(a,b);
  return 0;
}
void jia(long &a,long &b,long ax,long bx)
{

  a=a*bx+ax*b;
  b=bx*b;
  int i=2;
  while(i<=abs(a)&&i<=abs(b))
  {
    if(a%i==0&&b%i==0)
    {
      a=a/i;
      b=b/i;
      i=2;
      continue;
    }
    i++;
  }
}
void out(long a,long b)
{
  if(a*b<0&&b<0)
  {
    a=-a;
    b=-b;
  }
  if(b==1||a==0)
    cout<<a<<endl;
  else
  {
    int p=0;
    while(abs(a)>abs(b))
    {
      if(a*b>0)
      {
        p++;
        a=a-b;
      }
      else
      {
        p--;
        a=a+b;
      }
    }
    if(p!=0)
      cout<<p<<" ";
    if(a<0)
      a=-a;
    cout<<a<<"/"<<b<<endl;
  }
}