1081. Rational Sum (20) -最大公约数

题目例如以下：

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

题目要求对分数进行处理。题目的关键在于求取最大公约数，最初我採用了循环出现超时，后来改用辗转相除法。攻克了此问题。

须要注意的是分子为负数的情况，为方便处理，我们把负数取绝对值，而且记录下符号。最后再输出。

辗转相除法例如以下：

给定数a、b，要求他们的最大公约数，用随意一个除以还有一个，得到余数c，假设c=0，则说明除尽。除数就是最大公约数；假设c≠0，则用除数再去除以余数，如此循环下去，直至c=0，则除数就是最大公约数，直接说比較抽象，以下用样例说明。

设a=25，b=10。c为余数

①25/10，c=5≠0，令a=10。b=5。

②10/5。c=0，则b=5就是最大公约数。

求取最大公约数的代码例如以下：

long getMaxCommon(long a, long b){
    long yu;
    if(a == b) return a;
    while(1){
        yu = a % b;
        if(yu == 0) return b;
        a = b;
        b = yu;
    }
}

完整代码例如以下：

#include <iostream>
#include <stdio.h>
#include <vector>

using namespace std;

struct Ration{
    long num;
    long den;

    Ration(long _n, long _d){
        num = _n;
        den = _d;
    }

};

long getMaxCommon(long a, long b){
    long yu;
    if(a == b) return a;
    while(1){
        yu = a % b;
        if(yu == 0) return b;
        a = b;
        b = yu;
    }
}

int main(){
    int N;
    long num,den;
    long maxDen = -1;
    cin >> N;
    vector<Ration> rations;
    for(int i = 0; i < N; i++){
        scanf("%ld/%ld",&num,&den);
        rations.push_back(Ration(num,den));
        if(maxDen == -1){
            maxDen = den;
        }else{
            // 找maxDen和当前的最小公倍数
            if(den == maxDen) continue;
            else if(maxDen > den){
                if(maxDen % den == 0) continue;
            }else{
                if(den % maxDen == 0){
                    maxDen = den;
                    continue;
                }
            }
            maxDen = maxDen * den;
        }
    }
    num = 0;
    for(int i = 0; i < N; i++){
        num += rations[i].num * (maxDen / rations[i].den);
    }
    if(num == 0) {
        printf("0\n");
        return 0;
    }
    bool negative = num < 0;
    if(negative) num = -num;
    if(num >= maxDen){
        long integer = num / maxDen;
        long numerator = num % maxDen;
        if(numerator == 0){
            if(negative)
                printf("-%ld\n",integer);
            else
                printf("%ld\n",integer);
            return 0;
        }
        long common = getMaxCommon(numerator,maxDen);
        if(negative){
            printf("%ld -%ld/%ld\n",integer,numerator/common,maxDen / common);
        }else{
            printf("%ld %ld/%ld\n",integer,numerator/common,maxDen / common);
        }
    }else{
        long common = getMaxCommon(num,maxDen);
        if(negative)
            printf("-%ld/%ld\n",num/common,maxDen/common);
        else
            printf("%ld/%ld\n",num/common,maxDen/common);
    }
    return 0;
}