给定一个由N 个整数组成的整数数组arr[] ，任务是为数组中的每个元素找到数组中最近的素数。如果数组不包含任何质数，则打印-1 。
例子：
Input: arr[] = {1, 2, 3, 1, 6}
Output: 2 2 3 3 3
Explanation:
For the subarray {1, 2}, the nearest prime number is 2.
For the subarray {3, 1, 6}, the nearest prime number is 3.
Input: arr[] = {8, 7, 12, 15, 3, 11}
Output: 7 7 7 3 3 11
Explanation:
For the subarray {8, 7, 12}, the nearest prime number is 7.
For the subarray {15, 3}, the nearest prime number is 3.
For the subarray {11}, the nearest prime number is 11 itself.
方法：
请按照以下步骤解决问题：
- 查找数组中的最大元素maxm 。
- 使用埃拉托色尼筛法计算并存储最大为maxm 的所有质数
- 遍历数组并存储素数的索引。
- 如果数组中不存在素数，则为所有索引打印 -1。
- 将curr指向由素数组成的第一个索引。
- 对于curr 之前的每个索引，打印arr[primes[curr]]作为最近的素数。
- 对于超过curr 的索引，将距离与 primes[curr] 和 primes[curr + 1] 进行比较。如果 primes[curr] 更接近，则打印arr[primes[curr]] 。否则，增加curr nad print arr[primes[curr]] 。
- 如果curr是数组中的最后一个素数，则为所有索引打印arr[primes[curr]] 。
下面是上述方法的实现：
C++
// C++ program to find nearest
// prime number in the array
// for all array elements
#include
using namespace std;
#define max 10000000
// Create a boolean array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
bool prime[max] = { false };
// Sieve of Eratosthenes function
void SieveOfEratosthenes(int maxm)
{
prime[0] = prime[1] = true;
for (int i = 2; i * i <= maxm; i++) {
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i and are
// less than i^2 are already been marked.
if (!prime[i]) {
for (int j = i * i; j <= maxm; j += i) {
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
void print_nearest_prime(int arr[], int N)
{
int maxm = *max_element(arr, arr + N);
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
vector primes;
for (int i = 0; i < N; i++) {
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.push_back(i);
}
// If no primes are present
// in the array
if (primes.size() == 0) {
for (int i = 0; i < N; i++) {
cout << -1 << " ";
}
return;
}
// Store the current prime
int curr = 0;
for (int i = 0; i < N; i++) {
// If the no further
// primes exist in the array
if (curr == primes.size() - 1
// For all indices less than
// that of the current prime
|| i <= primes[curr]) {
cout << arr[primes[curr]] << " ";
continue;
}
// If the current prime is
// nearer
if (abs(primes[curr] - i)
< abs(primes[curr + 1] - i)) {
cout << arr[primes[curr]] << " ";
}
// If the next prime is nearer
else {
// Make the next prime
// as the current
curr++;
cout << arr[primes[curr]] << " ";
}
}
}
// Driver Program
int main()
{
int N = 6;
int arr[] = { 8, 7, 12, 15, 3, 11 };
print_nearest_prime(arr, N);
return 0;
} Java
// Java program to find nearest
// prime number in the array
// for all array elements
import java.util.*;
class GFG{
static final int max = 10000000;
// Create a boolean array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
static boolean []prime = new boolean[max];
// Sieve of Eratosthenes function
static void SieveOfEratosthenes(int maxm)
{
prime[0] = prime[1] = true;
for(int i = 2; i * i <= maxm; i++)
{
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i
// and are less than i^2 are already
// been marked.
if (!prime[i])
{
for(int j = i * i;
j <= maxm; j += i)
{
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
static void print_nearest_prime(int arr[], int N)
{
int maxm = Arrays.stream(arr).max().getAsInt();
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
Vector primes = new Vector();
for(int i = 0; i < N; i++)
{
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.add(i);
}
// If no primes are present
// in the array
if (primes.size() == 0)
{
for(int i = 0; i < N; i++)
{
System.out.print(-1 + " ");
}
return;
}
// Store the current prime
int curr = 0;
for(int i = 0; i < N; i++)
{
// If the no further
// primes exist in the array
if (curr == primes.size() - 1 ||
// For all indices less than
// that of the current prime
i <= primes.get(curr))
{
System.out.print(
arr[primes.get(curr)] + " ");
continue;
}
// If the current prime is
// nearer
if (Math.abs(primes.get(curr) - i) <
Math.abs(primes.get(curr + 1) - i))
{
System.out.print(
arr[primes.get(curr)] + " ");
}
// If the next prime is nearer
else
{
// Make the next prime
// as the current
curr++;
System.out.print(
arr[primes.get(curr)] + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
int N = 6;
int arr[] = { 8, 7, 12, 15, 3, 11 };
print_nearest_prime(arr, N);
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 program to find nearest
# prime number in the array
# for all array elements
maxi = 10000000
# Create a boolean array and set all
# entries it as false. A value in
# prime[i] will be true if i is not a
# prime, else false
prime = [False] * (maxi)
# Sieve of Eratosthenes function
def SieveOfEratosthenes(maxm):
prime[0] = prime[1] = True
for i in range(2, maxm + 1):
if i * i > maxm:
break
# Update all multiples of i greater
# than or equal to the square of it
# numbers which are multiple of i and are
# less than i^2 are already been marked.
if (not prime[i]):
for j in range(i * i, maxm + 1, i):
prime[j] = True
# Function to find nearest
# prime number for all elements
def print_nearest_prime(arr, N):
maxm = max(arr)
# Compute and store all prime
# numbers up to maxm
SieveOfEratosthenes(maxm)
primes = []
for i in range(N):
# Store the indices of
# all primes
if (not prime[arr[i]]):
primes.append(i)
# If no primes are present
# in the array
if len(primes) == 0:
for i in range(N):
print(-1, end = " ")
return
# Store the current prime
curr = 0
for i in range(N):
# If the no further primes
# exist in the array
if (curr == len(primes) - 1 or
# For all indices less than
# that of the current prime
i <= primes[curr]):
print(arr[primes[curr]], end = " ")
continue
# If the current prime is
# nearer
if (abs(primes[curr] - i) <
abs(primes[curr + 1] - i)):
print(arr[primes[curr]], end = " ")
# If the next prime is nearer
else:
# Make the next prime
# as the current
curr += 1
print(arr[primes[curr]], end = " ")
# Driver code
if __name__ == '__main__':
N = 6
arr = [ 8, 7, 12, 15, 3, 11 ]
print_nearest_prime(arr, N)
# This code is contributed by mohit kumar 29C#
// C# program to find nearest
// prime number in the array
// for all array elements
using System;
using System.Linq;
using System.Collections.Generic;
class GFG{
static readonly int max = 10000000;
// Create a bool array and set all
// entries it as false. A value in
// prime[i] will be true if i is not a
// prime, else false
static bool []prime = new bool[max];
// Sieve of Eratosthenes function
static void SieveOfEratosthenes(int maxm)
{
prime[0] = prime[1] = true;
for(int i = 2; i * i <= maxm; i++)
{
// Update all multiples of i greater
// than or equal to the square of it
// numbers which are multiple of i
// and are less than i^2 are already
// been marked.
if (!prime[i])
{
for(int j = i * i;
j <= maxm; j += i)
{
prime[j] = true;
}
}
}
}
// Function to find nearest
// prime number for all elements
static void print_nearest_prime(int []arr,
int N)
{
int maxm = arr.Max();
// Compute and store all prime
// numbers up to maxm
SieveOfEratosthenes(maxm);
List primes = new List();
for(int i = 0; i < N; i++)
{
// Store the indices of
// all primes
if (!prime[arr[i]])
primes.Add(i);
}
// If no primes are present
// in the array
if (primes.Count == 0)
{
for(int i = 0; i < N; i++)
{
Console.Write(-1 + " ");
}
return;
}
// Store the current prime
int curr = 0;
for(int i = 0; i < N; i++)
{
// If the no further
// primes exist in the array
if (curr == primes.Count - 1 ||
// For all indices less than
// that of the current prime
i <= primes[curr])
{
Console.Write(
arr[primes[curr]] + " ");
continue;
}
// If the current prime is
// nearer
if (Math.Abs(primes[curr] - i) <
Math.Abs(primes[curr + 1] - i))
{
Console.Write(
arr[primes[curr]] + " ");
}
// If the next prime is nearer
else
{
// Make the next prime
// as the current
curr++;
Console.Write(
arr[primes[curr]] + " ");
}
}
}
// Driver code
public static void Main(String[] args)
{
int N = 6;
int []arr = { 8, 7, 12, 15, 3, 11 };
print_nearest_prime(arr, N);
}
}
// This code is contributed by PrinciRaj1992 输出：
7 7 7 3 3 11
时间复杂度： O(maxm * (log(log(maxm))) + N)
辅助空间： O(N)
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