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 Nrational 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​​​ （分子/分母）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（小数） 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

思路：

简单的分数计算

参考代码：

 

#include
#include
using namespace std;

typedef long long LL;

LL gcd(LL a,LL b){
   return b == 0 ? a : gcd(b , a % b);
}

struct Fraction
{
   LL up, down;
};

Fraction reduction(Fraction result){//化简
   if(result.down < 0){//如果分母为负数
      result.up = -result.up;
      result.down = -result.down;
   }
   if(result.up == 0){//如果分子为0
      result.down = 1;
   }else{
      int d = gcd(abs(result.up),abs(result.down));//求最大公约数
      result.up /= d;//化简
      result.down /= d;
   }

   return result;
}

Fraction add(Fraction f1, Fraction f2){
   Fraction result;
   result.up = f1.up * f2.down + f2.up * f1.down;//分数和分子
   result.down = f1.down * f2.down;//分数和分母
   return reduction(result);//返回结果化简
}

void showResult(Fraction r){//展示结果
   reduction(r);
   if(r.down == 1)printf("%lld\n", r.up);//分母为1
   else if(abs(r.up) > r.down){//假分数
      printf("%lld %lld/%lld\n", r.up / r.down, abs(r.up) % r.down,  r.down);
   }else{//真分数
      printf("%lld/%lld\n", r.up, r.down);
   }
}

int main(){
   int n;
   Fraction sum, temp;
   //赋予初始值
   sum.up = 0;
   sum.down = 1;
   scanf("%d",&n);
   for (int i = 0; i < n; ++i)
   {
      scanf("%lld/%lld", &temp.up, &temp.down);
      sum = add(sum,temp);
   }
   showResult(sum);
   return 0;
}