找到具有最小绝对差的给定分数的最接近分数

给定三个整数x 、 y和n，任务是找到这样的一对整数a, b (1 <= b <= n; 0 <= a)使得值|x/y – a/b|尽可能小，其中 |x|表示 x 的绝对值。

例子：

Input: x = 3, y = 7, n = 6
Output: 2/5
Explanation: a=2 and b=5 and b<=6 then 3/7 – 2/5 = 1/35, which is the minimum difference possible

Input: x = 12, y = 37, n = 5
Output: 1/3

方法：迭代分母。设分母为i 。那么需要选择这样一个分子d使得|x/y – d/i|是最小的。 d = (x*i)/y是一个不错的选择。还要检查d+1 。将答案从A/B更新为d/i 时，检查x/y – d/i < x/y -A/B或(B*xy*A) * (i*y) > (i*xy* d) * (B*y)。请按照以下步骤解决问题：

• 将变量A和B初始化为-1以存储答案。
• 使用变量i迭代范围[1, N]并执行以下步骤：
  • 将变量d初始化为(i*x)/y作为最接近的可能分子。
  • 如果d大于等于0且A等于-1或ABS(B * x – y * A) * ABS(i * y)大于ABS(i * x – y * d) * ABS(B * y)，然后将A的值设置为d ，将B的值设置为i。
  • 将d的值增加1。
  • 如果d大于等于0且A等于-1或ABS(B * x – y * A) * ABS(i * y)大于ABS(i * x – y * d) * ABS(B * y)，然后将A的值设置为d ，将B的值设置为i。
• 执行上述步骤后，打印A/B的值作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the absolute
// value of x
long long ABS(long long x) { return max(x, -x); }
 
// Function to find the fraction with
// minimum absolute difference
void findFraction(long long x, long long y, long long n)
{
 
    // Initialize the answer variables
    long long A = -1, B = -1;
 
    // Iterate over the range
    for (long long i = 1; i <= n; i++) {
 
        // Nearest fraction
        long long d = (i * x) / y;
 
        // x/y - d/i < x/y -A/B
        //(B*x-y*A) * (i*y) > (i*x-y*d) * (B*y)
        if (d >= 0
            && (A == -1
                || ABS(B * x - y * A) * ABS(i * y)
                       > ABS(i * x - y * d) * ABS(B * y)))
            A = d, B = i;
 
        // Check for d+1
        d++;
        if (d >= 0
            && (A == -1
                || ABS(B * x - y * A) * ABS(i * y)
                       > ABS(i * x - y * d) * ABS(B * y)))
            A = d, B = i;
    }
 
    // Print the answer
    cout << A << "/" << B << endl;
}
 
// Driver Code
int main()
{
 
    long long x = 3, y = 7, n = 6;
 
    findFraction(x, y, n);
 
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG
{
   
    // Function to find the absolute
    // value of x
    static long ABS(long x) { return Math.max(x, -x); }
 
    // Function to find the fraction with
    // minimum absolute difference
    static void findFraction(long x, long y, long n)
    {
 
        // Initialize the answer variables
        long A = -1, B = -1;
 
        // Iterate over the range
        for (long i = 1; i <= n; i++) {
 
            // Nearest fraction
            long d = (i * x) / y;
 
            // x/y - d/i < x/y -A/B
            //(B*x-y*A) * (i*y) > (i*x-y*d) * (B*y)
            if (d >= 0
                && (A == -1
                    || ABS(B * x - y * A) * ABS(i * y)
                           > ABS(i * x - y * d)
                                 * ABS(B * y)))
                A = d;
            B = i;
 
            // Check for d+1
            d++;
            if (d >= 0
                && (A == -1
                    || ABS(B * x - y * A) * ABS(i * y)
                           > ABS(i * x - y * d)
                                 * ABS(B * y)))
                A = d;
            B = i;
        }
        A--;
        B--;
       
        // Print the answer
        System.out.println(A + "/" + B);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        long x = 3;
        long y = 7;
        long n = 6;
 
        findFraction(x, y, n);
    }
}
 
// This code is contributed by lokeshpotta.

Python3

# Python3 program for the above approach
 
# Function to find the absolute
# value of x
def ABS(x):
     
    return max(x, -x)
 
# Function to find the fraction with
# minimum absolute difference
def findFraction(x, y, n):
 
    # Initialize the answer variables
    A = -1
    B = -1
 
    # Iterate over the range
    for i in range(1, n + 1):
 
        # Nearest fraction
        d = (i * x) // y
 
        # x/y - d/i < x/y -A/B
        # (B*x-y*A) * (i*y) > (i*x-y*d) * (B*y)
        if (d >= 0 and (A == -1 or
           ABS(B * x - y * A) * ABS(i * y) >
           ABS(i * x - y * d) * ABS(B * y))):
            A = d
            B = i
 
        # Check for d+1
        d += 1
        if (d >= 0 and (A == -1 or
           ABS(B * x - y * A) * ABS(i * y) >
           ABS(i * x - y * d) * ABS(B * y))):
            A = d
            B = i
 
    # Print the answer
    print(str(A) + "/" + str(B))
 
# Driver Code
if __name__ == '__main__':
 
    x = 3
    y = 7
    n = 6
 
    findFraction(x, y, n)
 
# This code is contributed by mohit kumar 29

C#

// C# code for the above approach
using System;
 
public class GFG{
 
    // Function to find the absolute
    // value of x
    static long ABS(long x) { return Math.Max(x, -x); }
 
    // Function to find the fraction with
    // minimum absolute difference
    static void findFraction(long x, long y, long n)
    {
 
        // Initialize the answer variables
        long A = -1, B = -1;
 
        // Iterate over the range
        for (long i = 1; i <= n; i++) {
 
            // Nearest fraction
            long d = (i * x) / y;
 
            // x/y - d/i < x/y -A/B
            //(B*x-y*A) * (i*y) > (i*x-y*d) * (B*y)
            if (d >= 0
                && (A == -1
                    || ABS(B * x - y * A) * ABS(i * y)
                           > ABS(i * x - y * d)
                                 * ABS(B * y)))
                A = d;
            B = i;
 
            // Check for d+1
            d++;
            if (d >= 0
                && (A == -1
                    || ABS(B * x - y * A) * ABS(i * y)
                           > ABS(i * x - y * d)
                                 * ABS(B * y)))
                A = d;
            B = i;
        }
        A--;
        B--;
       
        // Print the answer
        Console.Write(A + "/" + B);
    }
 
    // Driver Code
    static public void Main (){
 
        long x = 3;
        long y = 7;
        long n = 6;
 
        findFraction(x, y, n);
    }
}
 
// This code is contributed by shubhamsingh10.

Javascript

 
// JavaScript program for the above approach
 
// Function to find the absolute
// value of x
function ABS(x)
{
    return Math.max(x, -x);
}
 
// Function to find the fraction with
// minimum absolute difference
function findFraction(x, y, n)
{
     
    // Initialize the answer variables
    let A = -1, B = -1;
 
    // Iterate over the range
    for(let i = 1; i <= n; i++)
    {
         
        // Nearest fraction
        let d = Math.floor((i * x) / y);
 
        // x/y - d/i < x/y -A/B
        //(B*x-y*A) * (i*y) > (i*x-y*d) * (B*y)
        if (d >= 0 && (A == -1 || ABS(B * x - y * A) *
            ABS(i * y) > ABS(i * x - y * d) * ABS(B * y)))
            A = d;
             
        B = i;
 
        // Check for d+1
        d++;
        if (d >= 0 && (A == -1 || ABS(B * x - y * A) *
            ABS(i * y) > ABS(i * x - y * d) * ABS(B * y)))
            A = d;
             
        B = i;
    }
    A--;
    B--;
   
    // Print the answer
    document.write(A + "/" + B);
}
 
// Driver code
let x = 3;
let y = 7;
let n = 6;
 
findFraction(x, y, n);
 
// This code is contributed by sanjoy_62
 

输出

2/5

时间复杂度： O(N)
辅助空间： O(1)