1081 Rational Sum (20分)

文章目录

1 题目

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/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

2 解析

2.1 题意

输出多个分数的和。

2.2 思路

正常的分数运算。
long long保存值，以防溢出。
使用abs时，用algorithm文件里面的。

3 参考代码

#include 
#include 

using std::abs;

typedef long long ll;

struct Fraction{
    ll up;
    ll down;
}ans,a;

ll gcd(ll a, ll b){
    return !b ? a :gcd(b, a%b);
}

Fraction reduction(Fraction res){
    if(res.up < 0){
        res.up = -res.up;
        res.down = -res.down;
    }

    if(res.up == 0){
        res.down = 1;
    }else{
        ll d = gcd(res.up, res.down);
        res.up /= d;
        res.down /= d;
    }

    return  res;
}

Fraction add(Fraction f1, Fraction f2){
    Fraction res;
    res.up = f1.up * f2.down + f2.up * f1.down;
    res.down = f1.down * f2.down;
    return reduction(res);
}

void showResult(Fraction res){
    if(res.down == 1){
        printf("%lld", res.up);
    }else if(abs(res.up) > res.down){
        printf("%lld %lld/%lld", res.up/res.down, abs(res.up)%res.down, res.down);
    }else{
        printf("%lld/%lld", res.up, res.down);
    }
}

int main(int agrc, char const *agrv[]){
    int n;
    scanf("%d", &n);
    scanf("%lld/%lld", &ans.up, &ans.down);
    n--;
    while(n--){
        scanf("%lld/%lld", &a.up, &a.down);
        ans = add(ans, a);
    }
    showResult(ans);
    return 0;
}