PAT甲级刷题记录——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 (≤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/denominatorwhere integeris 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

思路

这题大体的思路比较简单，但是细节上的东西要注意的地方有很多。题目大意就是给你几个分数，让你计算这几个分数的和，如果是假分数，要输出带分数的形式。

下面讲一下这题的【注意点】：

• 因为给出的分数不一定是最简的（比如样例1的-2/60），因此，需要得到这个分数之后就对它进行化简，然后存入input数组里面；
• 化简的时候求分子分母的最大公约数需要取绝对值，即：gcd(abs(), abs())，而且化简函数里也要特判分子为0的情况；
• 为了防止经过乘法之后超出int型范围，分子分母最好采用long long型存储；
• 如果是假分数的形式，要输出带分数，即：整数+真分数，当真分数的分子为0时，不输出这个真分数，只输出一个整数；
• 同理，如果本就是真分数的形式，当真分数的分子为0时，这时只要输出0即可（不要输出0/1，这是最后一个测试点考察的内容）。

代码

#include
#include
#include
#include
#include
#include
using namespace std;
struct Fraction{
    long long up, down;
};
vector<Fraction> input;
int gcd(int a, int b){
    if(b==0) return a;
    else return gcd(b, a%b);
}
Fraction huajian(Fraction a){
    if(a.up==0){
        a.down = 1;
    }
    else{
        int yueshu = gcd(abs(a.up), abs(a.down));
        a.up /= yueshu;
        a.down /= yueshu;
    }
    return a;
}
Fraction add(Fraction a, Fraction b){
    Fraction c;
    c.up = a.up*b.down+b.up*a.down;
    c.down = a.down*b.down;
    return c;
}
int main()
{
    int N;
    cin>>N;
    for(int i=0;i<N;i++){
        string temp;
        cin>>temp;
        int pos = temp.find("/");
        string fenzi = temp.substr(0,pos);
        string fenmu = temp.substr(pos+1);
        Fraction tmp;
        tmp.up = stoll(fenzi);
        tmp.down = stoll(fenmu);
        tmp = huajian(tmp);
        input.push_back(tmp);
    }
    Fraction result;
    result.up = 0;
    result.down = 1;
    for(int i=0;i<input.size();i++){
        result = add(result, input[i]);
    }
    result = huajian(result);
    if(result.up>=result.down){//假分数
        if(abs(result.up)%result.down==0){
            cout<<result.up/result.down;//只有整数
        }
        else{
            cout<<result.up/result.down<<" "<<abs(result.up)%result.down<<"/"<<result.down;
            //带分数
        }
    }
    else{//真分数
        if(result.up==0) cout<<"0";
        else cout<<result.up<<"/"<<result.down;
    }
    return 0;
}