PAT 1081 Rational Sum

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

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

//  a/a1 + b/b1

ll to_int(string s){
    ll sum = ; bool flag = ;
    if(s[] == '-'){
        flag = ;
        s = s.substr(,s.size()-);
    }
    for(int i=;i < s.size();i++){
        sum = sum* + (s[i]-'');
    }
    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 = ,a1 = ;

    while(t--){
        string s; cin >> s;
        int pos = s.find('/');
        string stra = s.substr(,pos);
        string stra1 = s.substr(pos+,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("");

    return ;
}

——一开始把输出写错了。。