PAT A1081

1081. Rational Sum (20)

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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -1/6 1/8

Sample Output 3:

7/24

实现代码如下：

#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;ll gcd(ll a, ll b){    return b == 0 ? a : gcd(b, a % b);}struct Fraction{    ll up, down;};Fraction reduction(Fraction result){    if(result.down < 0){        result.up = - result.up;        result.down = - result.down;    }    if(result.up == 0){        result.down = 1;    } else {        int d = gcd(abs(result.up), abs(result.down));        result.up /= d;        result.down /= d;    }    return result;}Fraction add(Fraction f1, Fraction f2){    Fraction result;    result.up = f1.up * f2.down + f2.up * f1.down;    result.down = f1.down * f2.down;    return reduction(result);}void showResult(Fraction r){    reduction(r);    if(r.down == 1) printf("%lld\n", r.up);    else if(abs(r.up) > r.down){        printf("%lld %lld/%lld\n", r.up / r.down, abs(r.up) % r.down, r.down);    } else {        printf("%lld/%lld", r.up, r.down);    }}int main(){    int n;    scanf("%d", &n);    Fraction sum, temp;    sum.up = 0; sum.down = 1;    for(int i = 0; i < n; i++){        scanf("%lld/%lld\n", &temp.up, &temp.down);        sum = add(sum, temp);    }    showResult(sum);    return 0;}