hdu5912Fraction+式子計算

Problem Description

Mr. Frog recently studied how to add two fractions up, and he came up with an evil idea to trouble you by asking you to calculate the result of the formula below:

As a talent, can you figure out the answer correctly?

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains only one integer n (n≤8).

The second line contains n integers: a1,a2,⋯an(1≤ai≤10).

The third line contains n integers: b1,b2,⋯,bn(1≤bi≤10).

Output

For each case, print a line “Case #x: p q”, where x is the case number (starting from 1) and p/q indicates the answer.

You should promise that p/q is irreducible.

Sample Input

1

2

1 1

2 3

Sample Output

Case #1: 1 2

Hint

Here are the details for the first sample:

2/(1+3/1) = 1/2

Source

2016中國大學生程式設計競賽（長春）-重制賽

#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL a[],b[];
int main(){
    int t,n;
    cin>>t;
    for(int cas=;cas<=t;cas++){
        cin>>n;
        for(int i=;i<=n;i++) cin>>a[i];
        for(int i=;i<=n;i++) cin>>b[i];
        LL fenzi=b[n],fenmu=a[n];

        for(int i=n-;i>=;i--){
            fenzi+=a[i]*fenmu;
            fenmu*=b[i];
            LL tem=__gcd(fenzi,fenmu);
            fenzi/=tem;
            fenmu/=tem;
            swap(fenzi,fenmu);
        }
        LL tem=__gcd(fenzi,fenmu);
        fenzi/=tem;
        fenmu/=tem;
        cout<<"Case #"<<cas<<": "<<fenzi<<" "<<fenmu<<endl;
    }
    return ;
}