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;
}