给定一个数组 ARR []的长度为N，其值小于N时，所述任务是构造相同的长度的另一个数组A []，使得对于在所述阵列A []每第i个元素，ARR [i]是最后一个索引(1基于索引的)由A [i]的倍数组成。
例子：
Input: arr[] = {4, 1, 2, 3, 4}
Output: 2 3 5 7 2
Explanation:
A[0]: Last index which can contain a multiple of A[0] has to be A[arr[0]] = A[4].
A[1]: Last index which can contain a multiple of A[1] has to be A[arr[1]] = A[1].
A[2]: Last index which can contain a multiple of A[2] has to be A[arr[2]] = A[2].
A[3]: Last index which can contain a multiple of A[3] has to be A[arr[3]] = A[3].
A[4]: Last index which can contain a multiple of A[4] has to be A[arr[4]] = A[4].
Hence, in the final array, A[4] must be divisible by A[0] and A[1], A[2] and A[3] must not be divisible by any other array elements.
Hence, the array A[] = {2, 3, 5, 7, 2} satisfies the condition.
Input: arr[] = {0, 1, 2, 3, 4}
Output: 2 3 5 7 11
方法：想法是将质数作为数组元素放置在满足条件的所需索引中。请按照以下步骤解决问题：
- 使用Eratosthenes筛生成所有素数 并将其存储在另一个数组中。
- 用{0}初始化数组A []以存储所需的数组。
- 遍历数组arr []并执行以下步骤：
- 检查A [arr [i]]是否为非零值，但A [i]为0。如果确定为true，则分配A [i] = A [arr [i]]。
- 检查A [arr [i]]和A [i]是否均为0。如果发现为真，则将与已经分配的数组元素不同的质数分配给两个索引arr [i]和i 。
- 完成上述步骤后，打印数组A []的元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
int sieve[1000000];
// Function to generate all
// prime numbers upto 10^6
void sieveOfPrimes()
{
// Initialize sieve[] as 1
memset(sieve, 1, sizeof(sieve));
int N = 1000000;
// Iterate over the range [2, N]
for (int i = 2; i * i <= N; i++) {
// If current element is non-prime
if (sieve[i] == 0)
continue;
// Make all multiples of i as 0
for (int j = i * i; j <= N; j += i)
sieve[j] = 0;
}
}
// Function to construct an array A[]
// satisfying the given conditions
void getArray(int* arr, int N)
{
// Stores the resultant array
int A[N] = { 0 };
// Stores all prime numbers
vector v;
// Sieve of Erastosthenes
sieveOfPrimes();
for (int i = 2; i <= 1e5; i++)
// Append the integer i
// if it is a prime
if (sieve[i])
v.push_back(i);
// Indicates current position
// in list of prime numbers
int j = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
int ind = arr[i];
// If already filled with
// another prime number
if (A[i] != 0)
continue;
// If A[i] is not filled
// but A[ind] is filled
else if (A[ind] != 0)
// Store A[i] = A[ind]
A[i] = A[ind];
// If none of them were filled
else {
// To make sure A[i] does
// not affect other values,
// store next prime number
int prime = v[j++];
A[i] = prime;
A[ind] = A[i];
}
}
// Print the resultant array
for (int i = 0; i < N; i++) {
cout << A[i] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 4, 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
getArray(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
static int[] sieve = new int[10000000];
// Function to generate all
// prime numbers upto 10^6
static void sieveOfPrimes()
{
// Initialize sieve[] as 1
Arrays.fill(sieve, 1);
int N = 1000000;
// Iterate over the range [2, N]
for (int i = 2; i * i <= N; i++)
{
// If current element is non-prime
if (sieve[i] == 0)
continue;
// Make all multiples of i as 0
for (int j = i * i; j <= N; j += i)
sieve[j] = 0;
}
}
// Function to construct an array A[]
// satisfying the given conditions
static void getArray(int[] arr, int N)
{
// Stores the resultant array
int A[] = new int[N];
Arrays.fill(A, 0);
// Stores all prime numbers
ArrayList v
= new ArrayList();
// Sieve of Erastosthenes
sieveOfPrimes();
for (int i = 2; i <= 1000000; i++)
// Append the integer i
// if it is a prime
if (sieve[i] != 0)
v.add(i);
// Indicates current position
// in list of prime numbers
int j = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
int ind = arr[i];
// If already filled with
// another prime number
if (A[i] != 0)
continue;
// If A[i] is not filled
// but A[ind] is filled
else if (A[ind] != 0)
// Store A[i] = A[ind]
A[i] = A[ind];
// If none of them were filled
else {
// To make sure A[i] does
// not affect other values,
// store next prime number
int prime = v.get(j++);
A[i] = prime;
A[ind] = A[i];
}
}
// Print the resultant array
for (int i = 0; i < N; i++) {
System.out.print( A[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 4, 1, 2, 3, 4 };
int N = arr.length;
// Function Call
getArray(arr, N);
}
}
// This code is contributed by code_hunt. Python3
# Python3 program for the above approach
sieve = [1]*(1000000+1)
# Function to generate all
# prime numbers upto 10^6
def sieveOfPrimes():
global sieve
N = 1000000
# Iterate over the range [2, N]
for i in range(2, N + 1):
if i * i > N:
break
# If current element is non-prime
if (sieve[i] == 0):
continue
# Make all multiples of i as 0
for j in range(i * i, N + 1, i):
sieve[j] = 0
# Function to construct an array A[]
# satisfying the given conditions
def getArray(arr, N):
global sieve
# Stores the resultant array
A = [0]*N
# Stores all prime numbers
v = []
# Sieve of Erastosthenes
sieveOfPrimes()
for i in range(2,int(1e5)+1):
# Append the integer i
# if it is a prime
if (sieve[i]):
v.append(i)
# Indicates current position
# in list of prime numbers
j = 0
# Traverse the array arr[]
for i in range(N):
ind = arr[i]
# If already filled with
# another prime number
if (A[i] != 0):
continue
# If A[i] is not filled
# but A[ind] is filled
elif (A[ind] != 0):
# Store A[i] = A[ind]
A[i] = A[ind]
# If none of them were filled
else:
# To make sure A[i] does
# not affect other values,
# store next prime number
prime = v[j]
A[i] = prime
A[ind] = A[i]
j += 1
# Prthe resultant array
for i in range(N):
print(A[i], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [4, 1, 2, 3, 4]
N = len(arr)
# Function Call
getArray(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
static int[] sieve = new int[10000000];
// Function to generate all
// prime numbers upto 10^6
static void sieveOfPrimes()
{
// Initialize sieve[] as 1
for(int i = 0; i < 10000000; i++)
{
sieve[i] = 1;
}
int N = 1000000;
// Iterate over the range [2, N]
for (int i = 2; i * i <= N; i++)
{
// If current element is non-prime
if (sieve[i] == 0)
continue;
// Make all multiples of i as 0
for (int j = i * i; j <= N; j += i)
sieve[j] = 0;
}
}
// Function to construct an array A[]
// satisfying the given conditions
static void getArray(int[] arr, int N)
{
// Stores the resultant array
int[] A = new int[N];
for(int i = 0; i < N; i++)
{
A[i] = 0;
}
// Stores all prime numbers
List v
= new List();
// Sieve of Erastosthenes
sieveOfPrimes();
for (int i = 2; i <= 1000000; i++)
// Append the integer i
// if it is a prime
if (sieve[i] != 0)
v.Add(i);
// Indicates current position
// in list of prime numbers
int j = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
int ind = arr[i];
// If already filled with
// another prime number
if (A[i] != 0)
continue;
// If A[i] is not filled
// but A[ind] is filled
else if (A[ind] != 0)
// Store A[i] = A[ind]
A[i] = A[ind];
// If none of them were filled
else {
// To make sure A[i] does
// not affect other values,
// store next prime number
int prime = v[j++];
A[i] = prime;
A[ind] = A[i];
}
}
// Print the resultant array
for (int i = 0; i < N; i++)
{
Console.Write( A[i] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 4, 1, 2, 3, 4 };
int N = arr.Length;
// Function Call
getArray(arr, N);
}
}
// This code is contributed by splevel62. 输出：
2 3 5 7 2时间复杂度： O(N * log(log(N)))
辅助空间： O(N)