pta 天梯赛 L1-009 N个数求和 c++题解

解题思路：通分，相加，约分求解

#include
#include
using namespace std;
int n, up, down, sumup, sumdown = 1;//需要对分母进行初始化
char c;
int gcd(int a,int b) 
{
    return b ? gcd(b, a % b) : a;
}
int main()
{
    cin >> n;
    while(n --)
    {
        cin >> up >>c >> down;
        sumup = sumup * down + up * sumdown;//通分
        sumdown *= down;
        
        int g = abs(gcd(sumup, sumdown)); //找到最大公约数
        sumup /=g; //通分
        sumdown /=g;
    }
    
    //输出结果
    if(sumup % sumdown == 0) 
        cout << sumup/ sumdown << endl; //结果为整数时
    else
    {
        if(sumup / sumdown)
            cout << sumup/ sumdown << " ";//当分子大于分母时
        cout << sumup - sumup/sumdown * sumdown<< '/' << sumdown << endl;
    }
    return 0;
}