生成一个 N 长度数组，其和等于给定数组中相同索引元素的绝对差之和的两倍
原文:https://www . geesforgeks . org/generate-n-length-array，其和等于给定数组的相同索引元素的绝对差之和的两倍/
给定大小为 N 的阵列 arr[] ，任务是构建大小为 N 的阵列 brr[] ，满足以下条件:
- 在数组 brr[] 中的每对连续元素中，一个元素必须能被另一个元素整除，即 brr[i] 必须能被 brr[i + 1] 整除，反之亦然。
- 阵列中的每个I元素 brr[] 必须满足 brr[i] > = arr[i] / 2 。
- 数组 arr[] 元素的和必须大于或等于2 *σABS(arr[I]–brr[I])。
示例:
输入: arr[] = { 11，5，7，3，2 } T3】输出:8 4 4 2 T6】解释:T8】ABS(11–8)+ABS(5–4)+ABS(7–4)+ABS(3–2)+ABS(2–2)= 8 arr[0]+arr[1]+…+arr[4]= 28 2 * 8<= 28 和 因此，brr[]的可能值之一是 8 4 4 2 2。
输入: arr[] = { 11，7，5 } 输出: { 8，4，4 }
方法:这个想法基于以下观察:
如果 brr[i] 是 2 的最近次幂，并且小于或等于 arr[i] ，那么 brr[i] 必须大于或等于 arr[i] / 2 以及数组的元素之和， arr[] 必须大于或等于2 *σABS(arr[I]–brr
按照以下步骤解决问题:
- 初始化一个数组 brr[] ，存储满足给定条件的元素。
- 穿越阵T2【arr】。对于每一个 i th 元素，找到最近的小于或等于arr【I】的 2 的幂，存入brr【I】。
- 最后，打印数组 brr[] 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct an array
// with given conditions
void constructArray(int arr[], int N)
{
int brr[N] = { 0 };
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
int K = log(arr[i]) / log(2);
// Stores closest power of 2
// less than or equal to arr[i]
int R = pow(2, K);
// Stores R into brr[i]
brr[i] = R;
}
// Print array elements
for (int i = 0; i < N; i++) {
cout << brr[i] << " ";
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 11, 5, 7, 3, 2 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
constructArray(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to conan array
// with given conditions
static void constructArray(int arr[], int N)
{
int brr[] = new int[N];
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
int K = (int)(Math.log(arr[i]) /
Math.log(2));
// Stores closest power of 2
// less than or equal to arr[i]
int R = (int)Math.pow(2, K);
// Stores R into brr[i]
brr[i] = R;
}
// Print array elements
for(int i = 0; i < N; i++)
{
System.out.print(brr[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 11, 5, 7, 3, 2 };
// Size of the array
int N = arr.length;
// Function Call
constructArray(arr, N);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program for the above approach
from math import log
# Function to construct an array
# with given conditions
def constructArray(arr, N):
brr = [0]*N
# Traverse the array arr[]
for i in range(N):
K = int(log(arr[i])/log(2))
# Stores closest power of 2
# less than or equal to arr[i]
R = pow(2, K)
# Stores R into brr[i]
brr[i] = R
# Prarray elements
for i in range(N):
print(brr[i], end = " ")
# Driver Code
if __name__ == '__main__':
# Given array
arr = [11, 5, 7, 3, 2]
# Size of the array
N = len(arr)
# Function Call
constructArray(arr, N)
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to construct an array
// with given conditions
static void constructArray(int []arr, int N)
{
int[] brr = new int[N];
Array.Clear(brr, 0, brr.Length);
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
int K = (int)(Math.Log(arr[i]) /
Math.Log(2));
// Stores closest power of 2
// less than or equal to arr[i]
int R = (int)Math.Pow(2, K);
// Stores R into brr[i]
brr[i] = R;
}
// Print array elements
for(int i = 0; i < N; i++)
{
Console.Write(brr[i] + " ");
}
}
// Driver Code
public static void Main()
{
// Given array
int []arr = { 11, 5, 7, 3, 2 };
// Size of the array
int N = arr.Length;
// Function Call
constructArray(arr, N);
}
}
// This code is contributed by SURENDRA_GANGWAR
java 描述语言
<script>
// JavaScript program for the above approach
// Function to conan array
// with given conditions
function constructArray(arr , N) {
var brr = Array(N).fill(0);
// Traverse the array arr
for (i = 0; i < N; i++) {
var K = parseInt( (Math.log(arr[i]) / Math.log(2)));
// Stores closest power of 2
// less than or equal to arr[i]
var R = parseInt( Math.pow(2, K));
// Stores R into brr[i]
brr[i] = R;
}
// Print array elements
for (i = 0; i < N; i++) {
document.write(brr[i] + " ");
}
}
// Driver Code
// Given array
var arr = [ 11, 5, 7, 3, 2 ];
// Size of the array
var N = arr.length;
// Function Call
constructArray(arr, N);
// This code is contributed by todaysgaurav
</script>
Output:
8 4 4 2 2
时间复杂度:O(N) T5辅助空间:** O(N)
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