PAT_甲级_1081

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1081 Rational Sum

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.

刚开始有个例子怎么过不去，后来发现是计算最小公倍数的函数返回类型用了int，没有用long long。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
using namespace std;
string temp_num;
long long gcd(long long a, long long b)
{
    if (b == 0) {
        return a;
    } else {
        return gcd(b, a % b);
    }
}

long long lcm(long long a, long long b, long long gcd_num)
{
    return a / gcd_num * b;
}

struct info{
    int fh;
    long long pre;
    long long post;
} arr[110];
int main()
{
    int num;
    cin >> num;
    for (int i = 0; i < num; i++) {
        cin >> temp_num;
        string pre;
        string post;
        long long post_xg;
        if (temp_num[0] == '-') {
            arr[i].fh = 0;
            post_xg = temp_num.find('/');
            pre = temp_num.substr(1, post_xg);
            post = temp_num.substr(post_xg + 1, int(temp_num.size()) - post_xg - 1);
        } else {
            arr[i].fh = 1;
            post_xg = temp_num.find('/');
            pre = temp_num.substr(0, post_xg);
            post = temp_num.substr(post_xg + 1, int(temp_num.size()) - post_xg - 1);
        }
        arr[i].pre = atoi(pre.c_str());
        arr[i].post = atoi(post.c_str());
    }
    long long temp = arr[0].post;
    for (int i = 1; i < num; i++) {
        long long temp_2 = gcd(arr[i].post, temp);
        temp = lcm(arr[i].post, temp, temp_2);
    }
    long long pre_num = 0;
    for (int i = 0; i < num; i++) {
        if (arr[i].fh == 1) {
            pre_num += temp / arr[i].post * arr[i].pre;
        } else {
            pre_num -= temp / arr[i].post * arr[i].pre;
        }
    }
    if (pre_num < 0) {
        cout << '-';
        pre_num = -pre_num;
    }
    if (pre_num == 0) {
        cout << '0' << endl;
        return 0;
    }
    int flag = 0;
    if (pre_num > temp) {
        long long r = pre_num / temp;
        cout << r;
        flag = 1;
    }
    if (pre_num % temp == 0) {
        return 0;
    }
    if (flag == 1) {
        cout << ' ';
    }
    long long rr = pre_num % temp;
    long long t = gcd(rr, temp);
    cout << rr / t << '/' << temp / t << endl;
    return 0;
}