1081. Rational Sum (20) -最大公约数

题目如下：

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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -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，最初我采用了循环出现超时，后来改用辗转相除法，解决了此问题。需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，并且记录下符号，最后再输出。

辗转相除法如下：

给定数a、b，要求他们的最大公约数，用任意一个除以另一个，得到余数c，如果c=0，则说明除尽，除数就是最大公约数；如果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，直接说比较抽象，下面用例子说明。

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如下：

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;    }}

完整代码如下：

#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;}