POH 4180 RealPhobia（连分数）

最新推荐文章于 2025-09-06 09:26:50 发布

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最新推荐文章于 2025-09-06 09:26:50 发布

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文章标签：人工智能 java

本文介绍了一个算法挑战，旨在帮助程序员Bert解决他对浮点运算的恐惧，通过将分数A/B转换为最接近的分数C/D，使得分母D小于B且误差最小。文章提供了完整的算法实现，包括如何分解为连分数并调整得到最优解。

RealPhobia

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 69 Accepted Submission(s): 27

Problem Description

Bert is a programmer with a real fear of floating point arithmetic. Bert has quite successfully used rational numbers to write his programs but he does not like it when the denominator grows large. Your task is to help Bert by writing a program that decreases the denominator of a rational number, whilst introducing the smallest error possible. For a rational number A/B, where B > 2 and 0 < A < B, your program needs to identify a rational number C/D such that:
1. 0 < C < D < B, and
2. the error |A/B - C/D| is the minimum over all possible values of C and D, and
3. D is the smallest such positive integer.

Input

The input starts with an integer K (1 <= K <= 1000) that represents the number of cases on a line by itself. Each of the following K lines describes one of the cases and consists of a fraction formatted as two integers, A and B, separated by “/” such that:
1. B is a 32 bit integer strictly greater than 2, and
2. 0 < A < B

Output

For each case, the output consists of a fraction on a line by itself. The fraction should be formatted as two integers separated by “/”.

Sample Input

   3 1/4 2/3 13/21
  

Sample Output

   1/3 1/2 8/13
  

Source

The 2011 South Pacific Programming Contest

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分解成连分数，然后最后一个数减一；

//HDU 4180 连分数 

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int an[100];
int gcd(int a,int b)
{
    int r=0;
    while(b!=0)
    {
        r=a%b;
        a=b;
        b=r;
    }    
    return a;
}    
int fenjie(int a,int b,int an[])//连分数分解 
{
    int n=0;
    int t;
    while(a!=1)
    {
        an[n++]=b/a;
        t=b%a;
        b=a;
        a=t;
    }    
    an[n++]=b;
    return n;
}    
int main()
{
    int n,A,B,C,D;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d/%d",&A,&B);
            int d=gcd(A,B);
            if(d>1)
            {
                printf("%d/%d\n",A/d,B/d);
                continue;
            }  
            int len=fenjie(A,B,an); 
            an[len-1]--;
            C=1;
            D=an[len-1];
            for(int j=len-2;j>=0;j--)
            {
                int t=an[j]*D+C;
                C=D;
                D=t;
            }    
            printf("%d/%d\n",C,D);
            
        }    
        
    }   
    return 0; 
}