PAT 甲级 1081 Rational Sum (20分)（多个分式相加，gcd忘了）*

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 (≤), 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

题意：给N个有理数（以分⼦/分⺟的形式给出），计算这N个数的总和，最后总和要以（整数 分⼦/分⺟）的形式给出～没整数就不给整数，分子是0，就不给分数

题解：

　　先根据分数加法的公式累加，后分离出整数部分和分数部分分⼦和分⺟都在⻓整型内，所以不能⽤int存储，否则有⼀个测试点不通过⼀开始⼀直是浮点错误，按理来说应该是出现了/0或者%0的情况，找了半天也不知道错在哪⾥，后来注意到应该在累加的时候考虑是否会超出long long的范围，所以在累加每⼀步之前进⾏分⼦分⺟的约分处理，然后就AC了～

     最小公倍数，最小公因数的gcd有点忘记了，回来要复习呀~

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a;
ll b;
int n;
ll gcd(ll x,ll y){
    return y == 0 ? x : gcd(y , x % y);
}
int main(){
    cin>>n;
    char c;
    ll fenzi=0;//初始化 
    ll fenmu=1;
    for(int i=1;i<=n;i++){
        cin>>a>>c>>b;
        if(b<0){//分母<0特殊处理 
            a=-1*a;
            b=-1*b;
        }
        ll gc=gcd(fenmu,b);
        fenzi = fenzi*b*gc + a*fenmu*gc;//先分子，再分母 
        fenmu=fenmu*b*gc;
        gc=gcd(fenmu,fenzi);//要通分，否则会爆ll浮点错误 
        fenmu/=gc;
        fenzi/=gc;
    }
    ll zheng = fenzi/fenmu;
    fenzi = fenzi%fenmu;
    if(fenzi==0) cout<<zheng;
    else{
        if(zheng==0){
            cout<<fenzi<<"/"<<fenmu;
        }else{
            cout<<zheng<<" ";
            cout<<fenzi<<"/"<<fenmu;
        }
    } 
    return 0;
}