2016中国大学生程序设计竞赛(ccpc 长春) Fraction【模拟】

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

 

题意：读入n，输入两行数，第一行是a数组的n个数,第二行是b数组的n个数，按照题目如图所示模拟计算过程，最后结果分数化为最简。

思路：模拟。最后分子分母除以它们最大公约数即最简。

#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)//循环模拟计算过程 
        {
            x = y*b[i--] + a[j--]*t;
            y = t;
            t = x; 
        }
        t = gcd(x,y);//求两数最大公约数 
        printf("Case #%d: %d %d
",++t2,x/t,y/t);
    }
    return 0;
}