ZJU-PAT 1081. Rational Sum (20)

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

模拟题

#pragma warning (disable:4786)
#include<iostream>
#include<iomanip>
#include<string>
#include<cstdio>
#include<algorithm>
#include<map>
#include<sstream>
using namespace std;

long long int sum,ans,sum1;
int n;
struct Node
{
    long long int numerator,denominator;
    bool flag;
};
Node TT[105];

long long int LCM(long long int a,long long int b)
{
    while(b!=0)
    {
        long long int r=a%b;
        a=b;
        b=r;
    }
    return a;
}

void Jisuan()
{
    ans=TT[0].numerator;
    sum=TT[0].denominator;

    for(int i=1;i<n;i++)
    {
        sum1=sum/LCM(sum,TT[i].denominator)*TT[i].denominator;

        long long int p=sum1/sum;
        long long int mm=sum1/TT[i].denominator;

        ans=ans*p;
        ans=ans+TT[i].numerator*mm;
        sum=sum1;
    }
    sum1=LCM(ans,sum);
    ans=ans/sum1;
    sum=sum/sum1;
}

int main()
{
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lld/%lld",&TT[i].numerator,&TT[i].denominator);

            if(TT[i].numerator<0) TT[i].flag=false;
        }
        Jisuan();
        if(abs(ans)>sum)
        {
            int ii=ans/sum;
            int tt=ans%sum;
            cout<<ii;
            if(tt) cout<<" "<<tt<<"/"<<sum;
            cout<<endl;
        }
        else
        {
            if(ans) cout<<ans<<"/"<<sum<<endl;
            else cout<<"0"<<endl;
        }
    }
    return 0;
}