題目：https://www.nowcoder.com/pat/1/problem/4311

題目描述：

Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum.

輸入描述：

Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in thenext line N rational numbers "a1/b1 a2/b2 ..." where all the numerators and denominators are in the range

輸出描述：

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

輸入例子：

5

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

輸出例子：

3 1/3

解題思路：

這道題只要解決兩個分數求和問題就可以了

a / b + c / d = (a * d + b * c) / (b * d) 

先求分母的最小公倍數數，然後同分，分子相加，得到結果後，再約分，先算出整數部分，然後分子分母同除以最大公約數。

具體解題步驟：

定義結構體儲存小數

struct Fraction
{
    long int fz;//分子
    long int fm;//分母
};
 

將數字從string轉成int,，逐位取出轉成int，再放到對應位置

long int strToInt(string s)
{
    long int n = 0;
    for(int i = s.length()-1; i >= 0; i--)
    {
        long int tmp = s[i] - '0';
        n += tmp*pow(10,s.length()-1-i);
    }
    return n;
}

其中，如果是常數字符串，可以直接呼叫atoi()函式

下面將分數字符串轉成結構體儲存，分數的正負只存在分子中，分母為正。

struct Fraction* strToFra(string s)
{
    struct Fraction* f = new struct Fraction;
    long int index = s.find('/');//找到/的下標
    if(s[0] == '-')//分數為負
    {
    //  cout<<"f"<<endl;
        string s1  = s.substr(1, index - 1);//分離出除符號位的分子的字串
        f->fz = -strToInt(s1);//將分子字串轉成int並加上符號
    }
    else//正數分子分離
    {
    //  cout<<"z"<<endl;
        string s2 = s.substr(0, index);
      f->fz = strToInt(s2);
    }
    //分母分離，只為正
    string s3 = s.substr(index+1, s.length() - index - 1);
    f->fm = strToInt(s3);
    return f;
}

下面是分數相加部分，分母先同分，然後分子相加，將分數字符串s和分數字符串ss相加

struct Fraction* sum = new struct Fraction;//結構體動態記憶體分配
sum = strToFra(s);//sum存s 
struct Fraction* f = new struct Fraction;
f = strToFra(ss); //f存ss
long int lcmfm = LCM(sum->fm, f->fm);//求分數sum分母和分數f分母的最小公倍數
long int sumfz = lcmfm/sum->fm*sum->fz + lcmfm/f->fm*f->fz;//通分
sum->fz = sumfz;//通分後分子作為sum分子結果
sum->fm = lcmfm;//分母最小公倍數作為sum分母結果
delete f;//釋放記憶體

long int beishu = 0;//化簡後分數的整數部分
if(abs(sum->fz) >= abs(sum->fm))//分子的絕對值比分母絕對值大（不加絕對值負數分數不會化簡，如下圖一）
{
    beishu = sum->fz/sum->fm;
    sum->fz -= sum->fm*beishu;//分子減去整數部分
}   
long int gc = abs(GCD(sum->fz, sum->fm));//求分子分母最大公約數的絕對值（不加絕對值，如果負分數，分子分母符號變反，如下圖二）
sum->fz /= gc;//分子分母同除最大公約數的絕對值，化簡
sum->fm /= gc;
if(beishu == 0)//輸出沒有整數部分的分數
{
    if(sum->fz == 0)//分子為0，直接輸出0
        cout<<0<<endl;
    else
        cout<<sum->fz<<"/"<<sum->fm<<endl;
}       
else//輸出有整數部分的分數
{
    if(sum->fz == 0)//分子為0，直接輸出整數部分
        cout<<beishu<<endl;
    else
        cout<<beishu<<" "<<sum->fz<<"/"<<sum->fm<<endl;
}

                      圖一                              圖二

下面是完整程式碼：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<cmath>
using namespace std;
 
 
struct Fraction
{
    long int fz;
    long int fm;
};
 
long int strToInt(string s)
{
    long int n = 0;
    for(int i = s.length()-1; i >= 0; i--)
    {
        long int tmp = s[i] - '0';
        n += tmp*pow(10,s.length()-1-i);
    }
    return n;
}
 
struct Fraction* strToFra(string s)
{
    struct Fraction* f = new struct Fraction;
    long int index = s.find('/');
    if(s[0] == '-')
    {
    //  cout<<"f"<<endl;
        string s1  = s.substr(1, index - 1);
        f->fz = -strToInt(s1);
    }
    else
    {
    //  cout<<"z"<<endl;
        string s2 = s.substr(0, index);
        f->fz = strToInt(s2);
    }
    string s3 = s.substr(index+1, s.length() - index - 1);
    f->fm = strToInt(s3);
    return f;
}
 
int GCD(long int x, long int y)
{
    while(y)
    {
        long int tmp = y;
        y = x % y;
        x = tmp;
    }
//  cout<<"x"<<x<<endl;
    return x;
}
 
int LCM(long int x, long int y)
{
    long int g = GCD(x,y);
    long int res = x*y/g;
//  cout<<"res"<<res<<endl;
    return res; 
}
 
int main() 
{
    long int num;
    cin>>num;
    string s;
    cin>>s;
    struct Fraction* sum = new struct Fraction;
    sum = strToFra(s); 
    for(int i = 0; i < num -1; i++)
    {
        string ss;
        cin>>ss;
        struct Fraction* f = new struct Fraction;
        f = strToFra(ss); 
        long int lcmfm = LCM(sum->fm, f->fm);
        long int sumfz = lcmfm/sum->fm*sum->fz + lcmfm/f->fm*f->fz;
        sum->fz = sumfz;
        sum->fm = lcmfm;
        delete f;
    }
    long int beishu = 0;
    if(abs(sum->fz) >= abs(sum->fm))
    {
        beishu = sum->fz/sum->fm;
        sum->fz -= sum->fm*beishu;
    }   
    long int gc = abs(GCD(sum->fz, sum->fm));
    sum->fz /= gc;
    sum->fm /= gc;
    if(beishu == 0)
    {
       if(sum->fz == 0)
           cout<<0<<endl;
        else
            cout<<sum->fz<<"/"<<sum->fm<<endl;
    }       
    else
    {
        if(sum->fz == 0)
           cout<<beishu<<endl;
        else
            cout<<beishu<<" "<<sum->fz<<"/"<<sum->fm<<endl;
    }
}