hdu 5912(迭代+gcd)

Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 97    Accepted Submission(s): 64

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:求图上的公式的结果.直接迭代求然后约分

#include <bits/stdc++.h>
using namespace std;
map<int ,int> value;
int a[],b[];
int gcd(int x,int y){
    return y==?x:gcd(y,x%y);
}
int main()
{
    int tcase,t=;
    scanf("%d",&tcase);
    while(tcase--){
        int n,res1,res2;
        scanf("%d",&n);
        for(int i=;i<=n;i++) scanf("%d",&a[i]);
        for(int i=;i<=n;i++) scanf("%d",&b[i]);
        res1 = a[n],res2 = b[n];
        for(int i=n-;i>=;i--){
            int k = res1;
            res1 = a[i]*res1+res2;
            res2 = b[i]*k;
        }
        int d = gcd(res1,res2);
        printf("Case #%d: %d %d\n",t++,res2/d,res1/d);
    }
    return ;
}