hdoj 5912 Fraction（模拟）

最新推荐文章于 2020-03-01 13:39:24 发布

cillyb

最新推荐文章于 2020-03-01 13:39:24 发布

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分类专栏：模拟水题文章标签： hdoj 模拟

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本文介绍了一种解决特殊分数累加问题的算法，通过逆向思维和模拟运算实现高效求解，适用于小规模数据集。

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题目地址：点击打开链接

因为数据量很小，直接倒着推模拟即可。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 15;
ll a[maxn], b[maxn], n;

int main(void)
{
    int ca = 1, t;
    cin >> t;
    while(t--)
    {
        scanf("%lld", &n);
        for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
        for(int i = 1; i <= n; i++) scanf("%lld", &b[i]);
        ll fz = b[n], fm = a[n];
        fz /= __gcd(a[n], b[n]);
        fm /= __gcd(a[n], b[n]);
        for(int i = n; i > 1; i--)
        {
            ll nfz = fm*b[i-1];
            ll nfm = fm*a[i-1]+fz;
            fz = nfz/__gcd(nfz, nfm);
            fm = nfm/__gcd(nfz, nfm);
        }
        printf("Case #%d: %lld %lld\n", ca++, fz, fm);
    }
    return 0;
}

Fraction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 836 Accepted Submission(s): 457

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

Source

2016中国大学生程序设计竞赛（长春）-重现赛