PAT甲级练习 Rational Sum (20)

题目描述
Given N rational numbers in the form “numerator/denominator”, you are supposed to calculate their sum.

输入描述:
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.

输出描述:
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.

#include<iostream>
using namespace std;
 
long int lcm(long int m, long int n)
{
 
    while(n != 0)
    {
        long int z = m % n; 
        m = n; 
        n = z; 
    }
    return m;
}
int main()
{
    int t;
    long int a[101], b[101], res;
    long int e, f;
    long int tempGcd1,tempLcm1, tempGcd2;
    long int p, q;
    long int u, v;
    int i;
    scanf("%d",&t);
    for(i = 0; i < t; i++)
    {
        scanf("%ld/%ld",&a[i],&b[i]);
    }
    p = b[0]; //分母1
    q = b[1]; //分母2
    u = a[0]; //分子1
    v = a[1]; //分子2
    for(i = 0; i < t-1; i++)
    {
        v = a[i+1];
        q = b[i+1];
        tempGcd1 = lcm(p,q); //最大公约数
 
        tempLcm1 = p * q / tempGcd1; //最小公倍数
 
        e = u * (tempLcm1 / p); //通分后分子1
 
        f = v *(tempLcm1 / q); //通分后分子2
 
        //printf("%ld/%ld ",e+f,tempLcm1);
         
        //对结果进行约分
        long int tempNumerator;
        if(e + f >= 0)
        {
        tempNumerator = e + f;
        }
        else
        {
            tempNumerator = -1*(e + f);
        }
 
        tempGcd2 = lcm(tempNumerator,tempLcm1);
 
        p = tempLcm1 / tempGcd2;
 
        u = (e + f) / tempGcd2;
        //printf("%ld/%ld ",u,p);
    }
    if(u == 0)
    {
        printf("%ld",0);
    }
 
    else if(u < p)
    {
        if(p == 1)
        {
            printf("%ld",u);
        }
        else
            printf("%ld/%ld",u, p);
    }
    else
    {
        res = u / p;
 
        if(u-res*p == 0)
        {
            printf("%ld",res);
        }
        else {
        printf("%ld %ld/%ld",res,u - res*p, p);
        }
    }
    return 0;
}