Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2    Accepted Submission(s): 2

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: b

1,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
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題目連結：

題目大意：

　　給你一個分式，如圖，求化簡後的分子分母(不含公約數)。

　　

題目思路：

　　【模擬】

　　因為n只有10，可以直接模擬。

//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define N 2004
using namespace std;
typedef long long LL;
double anss;
LL aans,sum;
int cas,cass;
int n,m,lll,ans;
int a[N],b[N];
int gcd(int a,int b)
{
	if(!b)return a;
	return gcd(b,a%b);
}
int main()
{
	#ifndef ONLINE_JUDGEW
//	freopen("1.txt","r",stdin);
//	freopen("2.txt","w",stdout);
	#endif
	int i,j,k;
//	init();
//	for(scanf("%d",&cass);cass;cass--)
	for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
//	while(~scanf("%s",s))
//	while(~scanf("%d",&n))
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)scanf("%d",&a[i]);
		for(i=1;i<=n;i++)scanf("%d",&b[i]);
		int fz=b[n],fm=a[n];
		for(i=n-1;i;i--)
		{
			fz+=a[i]*fm;
			fm*=b[i];
			swap(fz,fm);
		}
		i=gcd(fz,fm);
		fz/=i,fm/=i;
		printf("Case #%d: ",cass);
		printf("%d %d\n",fz,fm);
	}
	return 0;
}
/*
//

//
*/