HDU Fraction 2016中国大学生程序设计竞赛（长春）-重现赛-优快云博客

本文介绍了一个分数运算的编程挑战，任务是实现一个程序来计算特定分数公式的值，并确保结果为最简形式。输入包括测试案例数量及每组案例中分子与分母的数值，输出则是每个案例的计算结果。

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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

Sample Output

Case #1: 1 2

Hint
Here are the details for the first sample:
2/(1+3/1) = 1/2

水题直接暴力

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>
//#define CHECK_TIME

int a[11];
int b[11];

int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}

struct S{
    int fz,fm;
    S operator+(const S&a){
        S ans={fz*a.fm+a.fz*fm,fm*a.fm};
        ans.relax();
        return ans;
    }
    S operator -(const S&a){
        S ans={fz*a.fm-a.fz*fm,fm*a.fm};
        ans.relax();
        return ans;
    }
    S operator/(const S&a){
        S ans={fz*a.fm,fm*a.fz};
        ans.relax();
        return ans;
    }
    void relax(){
        int divi=gcd(fm,abs(fz));
        fm/=divi;
        fz/=divi;
    }
};

int main()
{
    //freopen("/home/lu/文档/r.txt","r",stdin);
    //freopen("/home/lu/文档/w.txt","w",stdout);
    int t,n;
    cin>>t;
    for(int tt=1;tt<=t;++tt){
        cin>>n;
        for(int i=1;i<=n;++i)
            cin>>a[i];
        for(int i=1;i<=n;++i)
            cin>>b[i];
        S ans={b[n],a[n]};
        ans.relax();
        for(int i=n-1;i>=1;--i){
            S ai={a[i],1};
            S bi={b[i],1};
            ans=bi/(ai+ans);
        }
        cout<<"Case #"<<tt<<": "<<ans.fz<<" "<<ans.fm<<endl;
    }
    return 0;
}