PAT Advanced 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 (≤), 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

在格式化的时候，要注意进行规整代码，乙级得考虑0这种特殊情况，防止浮点错误

#include <iostream>
#include <vector>
using namespace std;
struct frac{
    long long up, down, sign = 1;
};
long long gcd(long long a, long long b) {
    return b == 0 ? a: gcd(b, a%b);
}
int main() {
    long long N, multi = 1, sum = 0;
    scanf("%lld", &N);
    vector<frac> v;
    for(int i = 0; i < N; i++) {
        frac tmp;
        scanf("%lld/%lld", &tmp.up, &tmp.down);
        if(tmp.up < 0) {
            tmp.sign = -1;
            tmp.up = -tmp.up;
        }
        v.push_back(tmp);
        multi = multi * tmp.down;
    }
    for(int i = 0; i < N; i++)
        sum += (multi / v[i].down * v[i].up * v[i].sign);
    if(sum < 0) printf("-");
    if(sum == 0) {
        printf("0");
        return 0;
    }
    sum = abs(sum);
    int integer = sum / multi, blank = 0;
    sum %= multi;
    if(integer > 0) {
        printf("%d", integer);
        blank = true;
    }
    long long m = gcd(multi, sum);
    if(sum != 0) {
        if(blank) printf(" ");
        printf("%lld", sum/m);
        if(multi/m != 1L) printf("/%lld", multi/m);
    }
    return 0;
}