HDU 5912 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

 

2016中国大学生程序设计竞赛（长春）-重现赛

 

题目大意：给定a，b两个长度为n得数组，求出按图示公式计算后的分式的分子分母

思路: 模拟！由于n不大，可以直接从后往前模拟一边，最后求一遍最大公约数即可

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>

using namespace std;
int T,n;
int a[100],b[100];
int gcd(int c,int d){
    if(c==d)return c;
    else if(c<d)return gcd(d-c,c);
    return gcd(c-d,d);
}
void Swap(int &c,int &d){
    c = c^d;
    d = c^d;
    c = c^d;
}
int main()
{
    scanf("%d",&T);
    for(int t=1;t<=T;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]);
        int fa = a[n],fb = b[n];
        for(int i=n-1;i>=1;i--){
            fb = a[i]*fa+fb;
            Swap(fa,fb);
            fb *= b[i];
        }
        printf("Case #%d: ",t);
        int tmp = gcd(fa,fb);
        printf("%d %d
",fb/tmp,fa/tmp);
    }
    return 0;
}