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<vector>
#include<cstdio>
using namespace std;
struct Node{
	long long a;
	long long b;
};
long long find(long long x,long long y){
	long long tmp;
	if(x<y){
		tmp=x;
		x=y;
		y=tmp;
	}
	while(x%y!=0){
		tmp=x%y;
		x=y;
		y=tmp;
	}
	return y;
}
int main(){
	vector<Node>vt;
	int n;
	scanf("%d",&n);
	vt.resize(n);
	int i;
	for(i=0;i<n;i++){
		scanf("%lld/%lld",&vt[i].a,&vt[i].b);
	}
	long long a=0;
	long long b=1;
	for(i=0;i<n;i++){
		a=a*vt[i].b+b*vt[i].a;
		b=b*vt[i].b;
		long long tmp = find((a<0?-a:a),b);
		if(tmp==0)break;
		a=a/tmp;
		b=b/tmp;
	}
	long long m=a/b;
	long long k=a%b;
	if(m!=0&&k!=0){
		printf("%lld %lld/%lld\n",m,a-m*b,b);
	}else if(m==0&&k!=0){
		printf("%lld/%lld\n",a,b);
	}else if(m!=0&&k==0){
		printf("%lld\n",m);
	}else if(m==0&&k==0){
		printf("0\n");
	}
	return 0;
}