1.题目要求求一串分数相加的结果。
2.每次读取一个数时，对其进行因式化简，再累加，需要用到欧几里德算法。
3.每次直接进行相加，分子先不对分母取模，在最后才进行取模，才把整数部分提取出来。
4.注意在计算gcd时，取bot和top的绝对值，避免算出来的gcd为-1，使得top和bot的符号翻转。
5.刚开始尝试使用scanf(“%d/%d”,&a,&b)进行分式的读取，结果在读取负数分式的时候出错，所以改为读取string，然后再进行解析。
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
AC代码：
//#include<string>
//#include <iomanip>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
/*
3
1/3 -1/6 1/8
3
0/1 0/2 0/3
3
-1/2 -3/2 -1/3
4
-1/2 -3/2 -1/3 1/3
4
1/2 3/2 -1/3 1/3
4
1/2 3/2 -1/2 -3/2
*/
long long gcd(long long a, long long b)
{
return (b == 0 ? a : gcd(b, a%b));
}
struct ratNode{
long long top;
long long bot;
ratNode(int a, int b) :top(a), bot(b){};
ratNode() :top(0), bot(1){};
};
void add2Num(long long&nowTop, long long&nowBot, long long nextTop, long long nextBot)
{
long long botGCD = gcd(nowBot, nextBot);
long long a = nextBot / botGCD;
long long b = nowBot / botGCD;
long long top = nowTop*a + nextTop*b;
long long bot = a*nowBot;
long long GCD = gcd(labs(top), labs(bot));//注意取绝对值，避免算出来的gcd为-1，使得top和bot的符号翻转
nowTop = top / GCD;
nowBot = bot / GCD;
}
void str2num(string str, long long&top, long long&bot)
{
bool sign = true;
bool first = true;
top = 0; bot = 0;
for (int i = 0; i < str.size(); i++)
{
if (str[i] == '-')
sign = false;
else if (str[i] == '/')
first = false;
else if (first)
top = top * 10 + str[i] - '0';
else if (!first)
bot = bot * 10 + str[i] - '0';
}
if (!sign) top = -top;
}
int main(void)
{
int n;
cin >> n;
vector<ratNode> rat(n);
long long nowTop = 0;
long long nowBot = 1;
for (int i = 0; i < n; i++)
{
string str;
cin >> str;
str2num(str, rat[i].top, rat[i].bot);
if (rat[i].top == 0) rat[i].bot = 1;
int GCD = gcd(labs(rat[i].top), labs(rat[i].bot));//注意取绝对值，避免算出来的gcd为-1，使得top和bot的符号翻转
rat[i].top /= GCD;
rat[i].bot /= GCD;
add2Num(nowTop, nowBot, rat[i].top, rat[i].bot);
}
long long nowInt = nowTop / nowBot;
nowTop = nowTop - (nowInt*nowBot);
if (nowTop < 0 && nowInt!=0 ) nowTop = -nowTop;
if (nowTop == 0 && nowInt==0)
cout << "0" << endl;
else if (nowTop == 0)
cout << nowInt << endl;
else if (nowInt == 0)
cout << nowTop << "/" << nowBot << endl;
else
cout << nowInt << " " << nowTop << "/" << nowBot << endl;
return 0;
}