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📜  从给定列表中查找函数值最接近 A 的数字

📅  最后修改于: 2021-10-27 03:21:23             🧑  作者: Mango

给定一个函数F(n) = P – (0.006 * n) ,其中 P 是给定的。给定一个整数列表和一个数字, A .任务是从给定的列表中找到函数值最接近的数字A .
例子

Input : P = 12, A = 5
        List = {1000, 2000} 
Output : 1
Explanation :
Given, P=12, A=5
For 1000, F(1000) is 12 - 1000×0.006 = 6
For 2000, F(2000) is 12 - 2000×0.006 = 0
As the nearest value to 5 is 6, 
so the answer is 1000.

Input : P = 21, A = -11
        List = {81234, 94124, 52141}
Output : 3

方法:迭代给定列表中的每个值并为每个值找到 F(n)。现在,比较 F(n) 和 A 的每个值的绝对差与n ,其绝对差异最小就是答案。
下面是上述方法的实现:

C++
// C++ program to find number from
// given list for which value of the
// function is closest to A
#include 
using namespace std;
 
// Function to find number from
// given list for which value of the
// function is closest to A
int leastValue(int P, int A, int N, int a[])
{
 
    // Stores the final index
    int ans = -1;
 
    // Declaring a variable to store
    // the minimum absolute difference
    float tmp = (float)INFINITY;
 
    for (int i = 0; i < N; i++)
    {
 
        // Finding F(n)
        float t = P - a[i] * 0.006;
 
        // Updating the index of the answer if
        // new absolute difference is less than tmp
        if (abs(t-A) < tmp)
        {
            tmp = abs(t - A);
            ans = i;
        }
    }
    return a[ans];
}
 
// Driver code
int main()
{
    int N = 2, P = 12, A = 2005;
    int a[] = {1000, 2000};
 
    cout << leastValue(P, A, N, a) << endl;
}
 
// This code is contributed by
// sanjeev2552


Java
// Java program to find number from
// given list for which value of the
// function is closest to A
import java.util.*;
 
class GFG
{
 
// Function to find number from
// given list for which value of the
// function is closest to A
static int leastValue(int P, int A,
                      int N, int a[])
{
 
    // Stores the final index
    int ans = -1;
 
    // Declaring a variable to store
    // the minimum absolute difference
    float tmp = Float.MAX_VALUE;
 
    for (int i = 0; i < N; i++)
    {
 
        // Finding F(n)
        float t = (float) (P - a[i] * 0.006);
 
        // Updating the index of the answer if
        // new absolute difference is less than tmp
        if (Math.abs(t-A) < tmp)
        {
            tmp = Math.abs(t - A);
            ans = i;
        }
    }
    return a[ans];
}
 
// Driver code
public static void main(String[] args)
{
    int N = 2, P = 12, A = 2005;
    int a[] = {1000, 2000};
 
    System.out.println(leastValue(P, A, N, a));
}
}
 
// This code is contributed by 29AjayKumar


Python3
# Python program to find number from
# given list for which value of the
# function is closest to A
 
# Function to find number from
# given list for which value of the
# function is closest to A
def leastValue(P, A, N, a):
    # Stores the final index
    ans = -1
     
    # Declaring a variable to store
    # the minimum absolute difference
    tmp = float('inf')
    for i in range(N):
        # Finding F(n)
        t = P - a[i] * 0.006
         
        # Updating the index of the answer if
        # new absolute difference is less than tmp
        if abs(t - A) < tmp:
            tmp = abs(t - A)
            ans = i
             
    return a[ans]
 
# Driver Code
N, P, A = 2, 12, 5
a = [1000, 2000]
 
print(leastValue(P, A, N, a))


C#
// C# program to find number from
// given list for which value of the
// function is closest to A
using System;
     
class GFG
{
 
// Function to find number from
// given list for which value of the
// function is closest to A
static int leastValue(int P, int A,
                      int N, int []a)
{
 
    // Stores the final index
    int ans = -1;
 
    // Declaring a variable to store
    // the minimum absolute difference
    float tmp = float.MaxValue;
 
    for (int i = 0; i < N; i++)
    {
 
        // Finding F(n)
        float t = (float) (P - a[i] * 0.006);
 
        // Updating the index of the answer if
        // new absolute difference is less than tmp
        if (Math.Abs(t-A) < tmp)
        {
            tmp = Math.Abs(t - A);
            ans = i;
        }
    }
    return a[ans];
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 2, P = 12, A = 2005;
    int []a = {1000, 2000};
 
    Console.WriteLine(leastValue(P, A, N, a));
}
}
 
// This code is contributed by Rajput-Ji


PHP


Javascript


输出:
1000

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