# PAT A1081：Rational Sum之分数加法

1081 Rational Sum (20分)
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

``````#includecstdio
#includecmath
struct Fraction{
long long up,down;
};
int gcd(int a,int b)
{
return !b?a:gcd(b,a%b);
}
/*分数化简*/
Fraction reduction(Fraction result)
{
if(result.down0)	//分母为负数，令分子和分母都变为相反数
{
result.up=(-result.up);
result.down=(-result.down);
}
if(result.up==0)	//如果分子为0，则令分母为1
{
result.down=1;
}
else	//如果分子不为零，则进行约分，约去最大公约数
{
int d=gcd(abs(result.up),abs(result.down));
result.up/=d;
result.down/=d;
}
return result;
}
{
Fraction result;
result.up=f1.up*f2.down+f1.down*f2.up;
result.down=f1.down*f2.down;
return reduction(result);	//结果注意化简
}

void showResult(Fraction r)
{
reduction(r);
if(r.down==1)	printf("%lld\n",r.up);	//整数
else if(abs(r.up)r.down)
{
printf("%lld %lld/%lld\n",r.up/r.down,abs(r.up)%r.down,r.down);	//假分数
}
else
{
printf("%lld/%lld",r.up,r.down);	//真分数
}
}
int main()
{
int n;
scanf("%d",n);
Fraction sum,temp;
sum.up=0;
sum.down=1;
for(int i=0;in;i++)
{
scanf("%lld/%lld\n",temp.up,temp.down);
}
showResult(sum);
return 0;
}
``````
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