给定由N个正整数和两个正整数A和B组成的数组arr [] ，任务是用A和B的最接近幂的绝对差替换每个数组元素。如果存在两个最接近的幂，则选择两者中的最大值。
例子：
Input: arr[] = {5, 12, 25}, A = 2, B = 3
Output: 1 7 5
Explanation:
- For element arr[0]( = 5): The nearest power of 2 is 4 and the nearest power of 3 is 3. Therefore, the absolute difference of 4 and 3 is 1.
- For arr[1]( = 12): The nearest power of 2 is 16 and the nearest power of 3 is 9. Therefore, the absolute difference of 16 and 9 is 7.
- For arr[2]( = 5): The nearest power of 2 is 32 and the nearest power of 3 is 27. Therefore, the absolute difference of 27 and 32 is 5.
Therefore, the modified array is {1, 7, 5}.
Input: arr[] = {32, 3, 7}, a = 2, b = 3
Output: 0 1 1
方法：给定问题可以通过找到每个数组元素的A和B的最近幂并将其更新为获得的两个值的绝对差来解决。
请按照以下步骤解决问题
- 遍历给定数组arr []并执行以下步骤：
- 查找这两个值是一个小于和大于改编[I]完美的权力，并找到最接近的两个值。
- 找到两个小于b且大于arr [i]的b的完美幂的值，并找到两个值中最接近的值。
- 找到获得的两个最接近的值之间的绝对差，并使用它更新arr [i] 。
- 完成上述步骤后，打印修改后的数组arr [] 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the array
void printArray(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
void nearestPowerDiff(int arr[], int N,
int a, int b)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
// Find the log a of arr[i]
int log_a = log(arr[i]) / log(a);
// Find the power of a less
// than and greater than a
int A = pow(a, log_a);
int B = pow(a, log_a + 1);
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
int log_b = log(arr[i]) / log(b);
// Find the power of b less than
// and greater than b
A = pow(b, log_b);
B = pow(b, log_b + 1);
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver Code
int main()
{
int arr[] = { 5, 12, 25 };
int A = 2, B = 3;
int N = sizeof(arr) / sizeof(arr[0]);
nearestPowerDiff(arr, N, A, B);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to print the array
static void printArray(int[] arr, int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
System.out.print(arr[i] + " ");
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
static void nearestPowerDiff(int[] arr, int N,
int a, int b)
{
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Find the log a of arr[i]
int log_a = (int)(Math.log(arr[i]) /
Math.log(a));
// Find the power of a less
// than and greater than a
int A = (int)(Math.pow(a, log_a));
int B = (int)(Math.pow(a, log_a + 1));
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
int log_b = (int)(Math.log(arr[i]) /
Math.log(b));
// Find the power of b less than
// and greater than b
A = (int)(Math.pow(b, log_b));
B = (int)(Math.pow(b, log_b + 1));
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = Math.abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 5, 12, 25 };
int A = 2, B = 3;
int N = arr.length;
nearestPowerDiff(arr, N, A, B);
}
}
// This code is contributed by subham348Python3
# Python3 program for the above approach
import math
# Function to prthe array
def printArray(arr, N):
# Traverse the array
for i in range(N):
print(arr[i], end = " ")
# Function to modify array elements
# by absolute difference of the
# nearest perfect power of a and b
def nearestPowerDiff(arr, N, a, b):
# Traverse the array arr[]
for i in range(N):
# Find the log a of arr[i]
log_a = int(math.log(arr[i]) /
math.log(a))
# Find the power of a less
# than and greater than a
A = int(pow(a, log_a))
B = int(pow(a, log_a + 1))
if ((arr[i] - A) < (B - arr[i])):
log_a = A
else:
log_a = B
# Find the log b of arr[i]
log_b = int(math.log(arr[i]) /
math.log(b))
# Find the power of b less than
# and greater than b
A = int(pow(b, log_b))
B = int(pow(b, log_b + 1))
if ((arr[i] - A) < (B - arr[i])):
log_b = A
else:
log_b = B
# Update arr[i] with absolute
# difference of log_a & log _b
arr[i] = abs(log_a - log_b)
# Print the modified array
printArray(arr, N)
# Driver Code
arr = [ 5, 12, 25 ]
A = 2
B = 3
N = len(arr)
nearestPowerDiff(arr, N, A, B)
# This code is contributed by sanjoy_62C#
// C# program for the above approach
using System;
public class GFG{
// Function to print the array
static void printArray(int[] arr, int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
Console.Write(arr[i] + " ");
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
static void nearestPowerDiff(int[] arr, int N,
int a, int b)
{
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Find the log a of arr[i]
int log_a = (int)(Math.Log(arr[i]) /
Math.Log(a));
// Find the power of a less
// than and greater than a
int A = (int)(Math.Pow(a, log_a));
int B = (int)(Math.Pow(a, log_a + 1));
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
int log_b = (int)(Math.Log(arr[i]) /
Math.Log(b));
// Find the power of b less than
// and greater than b
A = (int)(Math.Pow(b, log_b));
B = (int)(Math.Pow(b, log_b + 1));
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = Math.Abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 5, 12, 25 };
int A = 2, B = 3;
int N = arr.Length;
nearestPowerDiff(arr, N, A, B);
}
}
// This code is contributed by AnkThon输出：
1 7 5时间复杂度： O(N)
辅助空间： O(1)