1081 Rational Sum （20 分）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
思路
　　先加再化简

#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#include<climits>
#include<sstream>
#include<cstdio>
#include<string.h>
#include<unordered_map>
using namespace std;
struct Fraction
{
    int up;
    int down;
};
int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
void reduction(Fraction &temp)
{
    if(temp.up==0)
        temp.down=1;
    int t=gcd(abs(temp.up),temp.down);
    temp.up/=t;
    temp.down/=t;
}

void add(Fraction &sum,Fraction &temp)
{
    sum.up=sum.up*temp.down+sum.down*temp.up;
    sum.down=sum.down*temp.down;
    reduction(sum);
}



int main()
{
    int n;
    scanf("%d",&n);
    Fraction sum,temp;
    for(int i=0;i<n;i++)
    {
        scanf("%d/%d",&temp.up,&temp.down);
        if(i==0)
            sum=temp;
        else
            add(sum,temp);
    }
    int it=sum.up/sum.down;
    if(it>0)
    {
        printf("%d",it);
        sum.up=sum.up%sum.down;
        if(sum.up!=0)
            printf(" %d/%d",sum.up,sum.down);
        return 0;
    }
    if(sum.up==0)
        printf("0");
    else
        printf("%d/%d",sum.up,sum.down);
    return 0;
}