PAT 1081 Rational Sum

1081 Rational Sum (20 分)

Given

Input Specification:

Each input file contains one test case. Each case starts with a positive integer

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:

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;


//  a/a1 + b/b1


ll to_int(string s){
    ll sum = 0; bool flag = 1;
    if(s[0] == '-'){
        flag = 0;
        s = s.substr(1,s.size()-1);
    }
    for(int i=0;i < s.size();i++){
        sum = sum*10 + (s[i]-'0');
    }
    if(flag) return sum;
    else return -sum;
}

ll gcd(ll x,ll y){
    return y?gcd(y,x%y):x;
}




//  a/a1 + b/b1

int main(){
    int t;cin >> t;
    ll a = 0,a1 = 1;

    while(t--){
        string s; cin >> s;
        int pos = s.find('/');
        string stra = s.substr(0,pos);
        string stra1 = s.substr(pos+1,s.size()-pos);
        ll b = to_int(stra);
        ll b1 = to_int(stra1);

        ll c = a*b1 + b*a1;
        ll c1 = a1*b1;

        ll t = gcd(c,c1);

        a = c/t;
        a1 = c1/t;
    }

    ll x = a/a1;
    ll y = a%a1;

    if(x&&y){printf("%lld %lld/%lld",x,y,a1);}
    else if(!x&&y){printf("%lld/%lld",y,a1);}
    else if(x&&!y)printf("%lld",x);
    else printf("0");

    return 0;
}

——一开始把输出写错了。。