skin 設計 中國大學 image tail action pre review 如圖所示

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 題意：讀入n，輸入兩行數，第一行是a數組的n個數,第二行是b數組的n個數，按照題目如圖所示模擬計算過程，最後結果分數化為最簡。 思路：模擬。最後分子分母除以它們最大公約數即最簡。

#include<stdio.h>
int
 
 a[20],b[20];

int gcd(int a,int b)
{
    int m;
    if( a < b)
        m = a,a=b,b=m;
    while(a%b)
    {
         m = a%b;
        a = b;
        b = m;
    }
    return b;
}

int main()
{
    int T,t,t2=0,i,n,j;
    int x,y;
    scanf("%d",&T);
    while(T --)
    {
        scanf("%d",&n);
        for(i = 1; i <= n; i ++)
            scanf("%d",&a[i]);
        for(i = 1; i <= n; i ++)
            scanf("%d",&b[i]);
        x = a[n-1]*a[n] + b[n];//初始化分子 
        y = a[n];//初始化分母 
        t = x;
        i = n-1;
        j = n-2;
        while(i>0||j>0)//循環模擬計算過程 
        {
            x = y*b[i--] + a[j--]*t;
            y = t;
            t = x; 
        }
        t = gcd(x,y);//求兩數最大公約數 
        printf("Case #%d: %d %d\n",++t2,x/t,y/t);
    }
    return 0;
}

2016中國大學生程序設計競賽(ccpc 長春) Fraction【模擬】