# 1081 Rational Sum （20 分）(数学题)

##### 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
``````
``````#include<iostream>
#include <cmath>
using namespace std;
int gcd(int a,int b) {
return b==0?a:gcd(b,a%b);
}
int main(){
int n;
cin>>n;
int a,b;
scanf("%d/%d",&a,&b);
int t=gcd(a,b);
a=a/t,b=b/t;
for(int i=1;i<n;i++){
int c,d;
scanf("%d/%d",&c,&d);
a=a*d+b*c;
b=b*d;
int t=gcd(a,b);
a=a/t,b=b/t;
}
if(a%b==0) printf("%d",a/b);
else{
if(abs(a)>abs(b)){
if(a>0) printf("%d %d/%d",a/b,a-a/b*b,b);
else{
a=-a;
printf("-%d %d/%d",a/b,a-a/b*b,b);
}
}else printf("%d/%d",a,b);
}
return 0;
}
``````
PAT