Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5912

题目大意：

　　给你一个分式，如图，求化简后的分子分母(不含公约数)。

题目思路：

　　【模拟】

　　因为n只有10，可以直接模拟。

 1 //
 2 //by coolxxx
 3 //#include<bits/stdc++.h>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<string>
 7 #include<iomanip>
 8 #include<map>
 9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #define min(a,b) ((a)<(b)?(a):(b))
21 #define max(a,b) ((a)>(b)?(a):(b))
22 #define abs(a) ((a)>0?(a):(-(a)))
23 #define lowbit(a) (a&(-a))
24 #define sqr(a) ((a)*(a))
25 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
26 #define mem(a,b) memset(a,b,sizeof(a))
27 #define eps (1e-10)
28 #define J 10
29 #define mod 1000000007
30 #define MAX 0x7f7f7f7f
31 #define PI 3.14159265358979323
32 #pragma comment(linker,"/STACK:1024000000,1024000000")
33 #define N 2004
34 using namespace std;
35 typedef long long LL;
36 double anss;
37 LL aans,sum;
38 int cas,cass;
39 int n,m,lll,ans;
40 int a[N],b[N];
41 int gcd(int a,int b)
42 {
43     if(!b)return a;
44     return gcd(b,a%b);
45 }
46 int main()
47 {
48     #ifndef ONLINE_JUDGEW
49 //    freopen("1.txt","r",stdin);
50 //    freopen("2.txt","w",stdout);
51     #endif
52     int i,j,k;
53 //    init();
54 //    for(scanf("%d",&cass);cass;cass--)
55     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
56 //    while(~scanf("%s",s))
57 //    while(~scanf("%d",&n))
58     {
59         scanf("%d",&n);
60         for(i=1;i<=n;i++)scanf("%d",&a[i]);
61         for(i=1;i<=n;i++)scanf("%d",&b[i]);
62         int fz=b[n],fm=a[n];
63         for(i=n-1;i;i--)
64         {
65             fz+=a[i]*fm;
66             fm*=b[i];
67             swap(fz,fm);
68         }
69         i=gcd(fz,fm);
70         fz/=i,fm/=i;
71         printf("Case #%d: ",cass);
72         printf("%d %d\n",fz,fm);
73     }
74     return 0;
75 }
76 /*
77 //
78
79 //
80 */