Best Rational Approximation( 法里数列)

Posted 2020-11-02 jadelemon

Best Rational Approximation

时间限制: 3 Sec  内存限制: 128 MB
提交: 297  解决: 25
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题目描述

Many microcontrollers have no floating point unit but do have a (reasonably) fast integer divide unit. In these cases it may pay to use rational values to approximate floating point constants. For instance,  
355/113 = 3.1415929203539823008849557522124  
is a quite good approximation to  
π = 3.14159265358979323846  
A best rational approximation, p/q, to a real number, x, with denominator at most M is a rational number, p/q (in lowest terms), with q <= M such that, for any integers, a and b with b <= M, and a and b relatively prime, p/q is at least as close to x as a/b:  
|x – p/q|≤|x – a/b|  
Write a program to compute the best rational approximation to a real number, x, with denominator at most M. 

输入

The first line of input contains a single integer P, (1≤P≤1000), which is the number of data sets that follow.  Each data set should be processed identically and independently.  
Each data set consists of a single line of input.  It contains the data set number, K, followed by the maximum denominator value, M (15≤M≤100000), followed by a floating-point value, x, (0≤x < 1). 

输出

For each data set there is a single line of output.  The single output line consists of the data set number, K, followed by a single space followed by the numerator, p, of the best rational approximation to x, followed by a forward slash (/) followed by the denominator, q, of the best rational approximation to x. 

样例输入

3
1 100000 .141592653589793238
2 255 .141592653589793238
3 15 .141592653589793238

样例输出

1 14093/99532
2 16/113
3 1/7

提示

 

来源

Greater New York Regional Contest 2017 

 

 

c++ code:

#include <bits/stdc++.h>

using namespace std;
int main()
{
    int p,t,m;
    double x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%lf",&p,&m,&x);
        int  a=0,b=1,c=1,d=1,e,f;
        while(true)
        {
            e=a+c;
            f=b+d;
            int gcd=__gcd(e,f);
            e/=gcd;f/=gcd;
            if(f>m) break;
            if(1.0*e/f<=x)
            {
                a=e;b=f;
            }
            else
            {
                c=e;d=f;
            }
        }
        printf("%d ",p);
        if(fabs(1.0*a/b-x)>fabs(1.0*c/d-x))
            printf("%d/%d\n",c,d);
        else
            printf("%d/%d\n",a,b);
    }
    return 0;
}