HDU5912-Fraction

Fraction

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                          Total Submission(s): 175    Accepted Submission(s): 111

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

121 12 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 <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;int gcd(int a,int b){    if(a>b) return gcd(b,a);    while(b%a)    {        int k=b%a;        b=a;        a=k;    }    return a;}int main(){    int t,n,cas=0,a[100],b[100],x,y;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        for(int i=1; i<=n; i++)            scanf("%d",&b[i]);        x=b[n],y=a[n];        for(int i=n-1; i>=1; i--)        {            x=y*a[i]+x;            int k=x;            x=y*b[i];            y=k;        }        int k=gcd(x,y);        printf("Case #%d: %d %d\n",++cas,x/k,y/k);    }    return 0;}