PAT Rational Sum

题目描述：

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

输入描述：

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 of "long int". If there is a negative number, then the sign must appear in front of the numerator.

输出描述：

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 common factor. You must output only the fractional part if the integer part is 0.

输入例子：

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

输出例子：

3 1/3

解题思路：

这道题只要解决两个分数求和问题就可以了

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

先求分母的最小公倍数数，然后同分，分子相加，得到结果后，再约分，先算出整数部分，然后分子分母同除以最大公约数。

具体解题步骤：

定义结构体存储小数

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

将数字从string转成int,，逐位取出转成int，再放到对应位置

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，如果是常数字符串，可以直接调用atoi()函数

下面将分数字符串转成结构体存储，分数的正负只存在分子中，分母为正。

struct Fraction* strToFra(string s)
{
    struct Fraction* f = new struct Fraction;
    long int index = s.find('/');//找到/的下标
    if(s[0] == '-')//分数为负
    {
    //  cout<<"f"<<endl;
        string s1  = s.substr(1, index - 1);//分离出除符号位的分子的字符串
        f->fz = -strToInt(s1);//将分子字符串转成int并加上符号
    }
    else//正数分子分离
    {
    //  cout<<"z"<<endl;
        string s2 = s.substr(0, index);

      f->fz = strToInt(s2);

}

    //分母分离，只为正
    string s3 = s.substr(index+1, s.length() - index - 1);
    f->fm = strToInt(s3);
    return f;
}

struct Fraction* strToFra(string s){ struct Fraction* f = new struct Fraction; long int index = s.find('/');//找到/的下标 if(s[0] == '-')//分数为负 { // cout<<"f"<<endl; string s1 = s.substr(1, index - 1);//分离出除符号位的分子的字符串 f->fz = -strToInt(s1);//将分子字符串转成int并加上符号 } else//正数分子 { // 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;}

下面是分数相加部分，分母先同分，然后分子相加，将分数字符串s和分数字符串ss相加

struct Fraction* sum = new struct Fraction;//结构体动态内存分配
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);//求分数sum分母和分数f分母的最小公倍数
long int sumfz = lcmfm/sum->fm*sum->fz + lcmfm/f->fm*f->fz;//通分
sum->fz = sumfz;//通分后分子作为sum分子结果
sum->fm = lcmfm;//分母最小公倍数作为sum分母结果
delete f;//释放内存

化简分数

long int beishu = 0;//化简后分数的整数部分
if(abs(sum->fz) >= abs(sum->fm))//分子的绝对值比分母绝对值大（不加绝对值负数分数不会化简，如下图一）
{
    beishu = sum->fz/sum->fm;
    sum->fz -= sum->fm*beishu;//分子减去整数部分
}   
long int gc = abs(GCD(sum->fz, sum->fm));//求分子分母最大公约数的绝对值（不加绝对值，如果负分数，分子分母符号变反，如下图二）
sum->fz /= gc;//分子分母同除最大公约数的绝对值，化简
sum->fm /= gc;
if(beishu == 0)//输出没有整数部分的分数
{
    if(sum->fz == 0)//分子为0，直接输出0
        cout<<0<<endl;
    else
        cout<<sum->fz<<"/"<<sum->fm<<endl;
}       
else//输出有整数部分的分数
{
    if(sum->fz == 0)//分子为0，直接输出整数部分
        cout<<beishu<<endl;
    else
        cout<<beishu<<" "<<sum->fz<<"/"<<sum->fm<<endl;
}

图一图二

下面是完整代码：

#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; }}