hdoj5972Fraction

Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 349    Accepted Submission(s): 204

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中国大学生程序设计竞赛（长春）-重现赛

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef pairpii;
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
int a[10],b[10];
int main()
{
    int t,Test=1;
    scanf("%d",&t);
    while(t--){
        int n;cin>>n;
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;++i){
            scanf("%d",&b[i]);
        }
        int ansa=b[n],ansb=a[n];
        for(int i=n-1;i>=1;--i){
            int temp=ansb;
            ansb=a[i]*ansb+ansa;
            ansa=b[i]*temp;

        }
        int g=gcd(ansa,ansb);
        printf("Case #%d: %d %d\n",Test++,ansa/g,ansb/g);
    }
    return 0;
}