检查给定的数字 K 是否足以到达数组的末尾
给定一个包含 n 个元素的数组 arr[] 和一个数字 K。任务是通过执行以下操作来确定是否有可能到达数组的末尾：
遍历给定的数组，并且，
- 如果发现任何元素不是素数，则将 K 的值减 1。
- 如果任何元素是素数，则将 K 的值重新填充为其初始值。
如果 (K > 0) 可以到达数组的末尾，则打印 YES，否则打印 NO。
例子：
Input : K = 2, arr[]={ 6, 3, 4, 5, 6};
Output : Yes
Explanation :
1- arr[0] is not prime, so K = K-1 = 1
2- arr[1] is prime so K will be refilled to its
initial value. Therefore, K = 2.
3- arr[2] is not prime.
Therefore, K = 2-1 = 1
4- arr[3] is prime so K will be refilled to its
initial value. Therefore, K = 2.
5- arr[4] is not prime.
Therefore, K = 2-1 = 1
6- Since the end of the array is reached with K>=0
So output is YES
Input : k=3, arr[]={ 1, 2, 10, 4, 6, 8};
Output : No推荐：请先在“练习”上解决，然后再继续解决。
简单的方法：
- 遍历数组的每个元素并检查当前元素的值是否为素数。
- 如果它是素数，则重新填充 K 的幂，否则减 1。
- 如果 (K > 0) 可以到达数组的末尾，则打印“YES”，否则打印“NO”。
下面是上述方法的实现：
C++
// C++ program to check if it is possible
// to reach the end of the array
#include
using namespace std;
// Function To check number is prime or not
bool is_Prime( int num )
{
// because 1 is not prime
if(num == 1)
return false;
for(int i=2 ; i*i <= num ; i++ )
{
if( num % i == 0 )
return false;
}
return true;
}
// Function to check whether it is possible
// to reach the end of the array or not
bool isReachable( int arr[] , int n , int k)
{
// store initial value of K
int x = k ;
for(int i=0 ; i < n ; i++ )
{
// Call is_prime function to
// check if a number is prime.
if( is_Prime(arr[i]) )
{
// Refill K to initial value
k = x;
}
else
{
// Decrement k by 1
k-- ;
}
if( k <= 0 && i < (n-1) && (!is_Prime(arr[i+1])) )
return false ;
}
return true ;
}
// Driver Code
int main()
{
int arr[] = { 6, 3, 4, 5, 6};
int n = sizeof(arr)/sizeof(arr[0]) ;
int k = 2 ;
isReachable( arr , n , k ) ? cout << "Yes" << endl :
cout << "No" << endl ;
return 0 ;
} Java
// Java program to check if
// it is possible to reach
// the end of the array
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG
{
// Function To check
// number is prime or not
static boolean is_Prime(int num)
{
// because 1 is not prime
if(num == 1)
return false;
for(int i = 2 ;
i * i <= num ; i++ )
{
if(num % i == 0)
return false;
}
return true;
}
// Function to check whether
// it is possible to reach
// the end of the array or not
static boolean isReachable(int arr[] ,
int n , int k)
{
// store initial value of K
int x = k ;
for(int i = 0 ; i < n ; i++ )
{
// Call is_prime function to
// check if a number is prime.
if(is_Prime(arr[i]))
{
// Refill K to
// initial value
k = x;
}
else
{
// Decrement k by 1
k-- ;
}
if(k <= 0 && i < (n - 1) &&
(is_Prime(arr[i + 1]) != true))
return false ;
}
return true ;
}
// Driver Code
public static void main(String args[])
{
int arr[] = new int[]{ 6, 3, 4, 5, 6};
int n = arr.length;
int k = 2 ;
if(isReachable(arr, n, k) == true)
System.out.print("Yes" + "\n");
else
System.out.print("No" + "\n");
}
}Python3
# Python 3 program to check if it is
# possible to reach the end of the array
from math import sqrt
# Function To check number is prime or not
def is_Prime(num):
# because 1 is not prime
if(num == 1):
return False
k = int(sqrt(num)) + 1
for i in range(2, k, 1):
if(num % i == 0):
return False
return True
# Function to check whether it is possible
# to reach the end of the array or not
def isReachable(arr, n , k):
# store initial value of K
x = k
for i in range(0, n, 1):
# Call is_prime function to
# check if a number is prime.
if( is_Prime(arr[i])):
# Refill K to initial value
k = x
else:
# Decrement k by 1
k -= 1
if(k <= 0 and i < (n - 1) and
(is_Prime(arr[i + 1])) == False):
return False
return True
# Driver Code
if __name__ == '__main__':
arr = [6, 3, 4, 5, 6]
n = len(arr)
k = 2
if (isReachable( arr , n , k )):
print("Yes")
else:
print("No")
# This code is contributed by
# Sahil_ShelangiaC#
// C# program to check if
// it is possible to reach
// the end of the array
using System;
class GFG
{
// Function To check
// number is prime or not
static bool is_Prime(int num)
{
// because 1 is not prime
if(num == 1)
return false;
for(int i = 2 ;
i * i <= num ; i++ )
{
if(num % i == 0)
return false;
}
return true;
}
// Function to check whether
// it is possible to reach
// the end of the array or not
static bool isReachable(int []arr ,
int n , int k)
{
// store initial
// value of K
int x = k ;
for(int i = 0 ; i < n ; i++ )
{
// Call is_prime function
// to check if a number
// is prime.
if(is_Prime(arr[i]))
{
// Refill K to
// initial value
k = x;
}
else
{
// Decrement k by 1
k-- ;
}
if(k <= 0 && i < (n - 1) &&
(is_Prime(arr[i + 1]) != true))
return false ;
}
return true ;
}
// Driver Code
public static void Main()
{
int []arr = new int[]{ 6, 3, 4, 5, 6};
int n = arr.Length;
int k = 2 ;
if(isReachable(arr, n, k) == true)
Console.WriteLine("Yes" + "\n");
else
Console.WriteLine("No" + "\n");
}
}
// This code is contributed by vt_mPHP
Javascript
C++
// C++ program to check if it is possible
// to reach the end of the array
#include
#define MAX 1000000
using namespace std;
// Function for Sieve of Eratosthenes
void SieveOfEratosthenes( int sieve[], int max )
{
for(int i=0; i Java
// Java program to check if it is possible
// to reach the end of the array
class GFG
{
static int MAX = 1000000;
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes(int sieve[], int max)
{
for (int i = 0; i < max; i++)
{
sieve[i] = 1;
}
for (int i = 2; i * i < max; i++)
{
if (sieve[i] == 1)
{
for (int j = i * 2; j < max; j += i)
{
sieve[j] = 0;
}
}
}
}
// Function to check if it is possible to
// reach end of the array
static boolean isReachable(int arr[], int n,
int sieve[], int k)
{
// store initial value of K
int x = k;
for (int i = 0; i < n; i++)
{
if (sieve[arr[i]] == 1)
{
// Refill K to initial value
k = x;
}
else
{
// Decrement k by 1
k -= 1;
}
if ((k <= 0) && (i < (n - 1))
&& (sieve[arr[i + 1]] == 0))
{
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {6, 3, 4, 5, 6};
int[] sieve = new int[MAX];
int n = arr.length;
int k = 2;
SieveOfEratosthenes(sieve, MAX);
if (isReachable(arr, n, sieve, k))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
/* This code contributed by PrinciRaj1992 */Python3
# Python3 program to check if it is
# possible to reach the end of the
# array
import math
# Function for Sieve of Eratosthenes
def SieveOfEratosthenes(sieve, max):
for i in range(0, max):
sieve[i] = 1
sqt = int(math.sqrt(max))
for i in range(2, sqt):
if (sieve[i] == 1):
for j in range(i * 2, max, i):
sieve[j] = 0
# Function to check if it is possible to
# reach end of the array
def isReachable(arr, n, sieve, k):
# store initial value of K
x = k
for i in range(0, n):
if (sieve[arr[i]] != 0):
# Refill K to initial value
k = x
else:
# Decrement k by 1
k -= 1
if ((k <= 0) and (i < (n - 1)) and
(sieve[arr[i + 1]] == 0)):
return 0
return 1
# Driver Code
arr = [ 6, 3, 4, 5, 6 ]
sieve = [0 for x in range(1000000)]
n = len(arr)
k = 2
SieveOfEratosthenes(sieve, 1000000)
ch = isReachable(arr, n, sieve, k)
if (ch):
print("Yes")
else:
print("No")
# This code is contributed by Stream_CipherC#
// C# program to check if it is possible
// to reach the end of the array
using System;
class GFG
{
static int MAX = 1000000;
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes(int []sieve, int max)
{
for (int i = 0; i < max; i++)
{
sieve[i] = 1;
}
for (int i = 2; i * i < max; i++)
{
if (sieve[i] == 1)
{
for (int j = i * 2; j < max; j += i)
{
sieve[j] = 0;
}
}
}
}
// Function to check if it is possible to
// reach end of the array
static bool isReachable(int []arr, int n,
int []sieve, int k)
{
// store initial value of K
int x = k;
for (int i = 0; i < n; i++)
{
if (sieve[arr[i]] == 1)
{
// Refill K to initial value
k = x;
}
else
{
// Decrement k by 1
k -= 1;
}
if ((k <= 0) && (i < (n - 1))
&& (sieve[arr[i + 1]] == 0))
{
return false;
}
}
return true;
}
// Driver Code
static public void Main ()
{
int []arr = {6, 3, 4, 5, 6};
int[] sieve = new int[MAX];
int n = arr.Length;
int k = 2;
SieveOfEratosthenes(sieve, MAX);
if (isReachable(arr, n, sieve, k))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
/* This code contributed by ajit. */Javascript
输出：
Yes时间复杂度： O(N(sqrt N))
有效的方法：上述方法可以通过使用埃拉托色尼筛来优化，以检查一个数字是否为素数。
下面是高效方法的实现：
C++
// C++ program to check if it is possible
// to reach the end of the array
#include
#define MAX 1000000
using namespace std;
// Function for Sieve of Eratosthenes
void SieveOfEratosthenes( int sieve[], int max )
{
for(int i=0; i Java
// Java program to check if it is possible
// to reach the end of the array
class GFG
{
static int MAX = 1000000;
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes(int sieve[], int max)
{
for (int i = 0; i < max; i++)
{
sieve[i] = 1;
}
for (int i = 2; i * i < max; i++)
{
if (sieve[i] == 1)
{
for (int j = i * 2; j < max; j += i)
{
sieve[j] = 0;
}
}
}
}
// Function to check if it is possible to
// reach end of the array
static boolean isReachable(int arr[], int n,
int sieve[], int k)
{
// store initial value of K
int x = k;
for (int i = 0; i < n; i++)
{
if (sieve[arr[i]] == 1)
{
// Refill K to initial value
k = x;
}
else
{
// Decrement k by 1
k -= 1;
}
if ((k <= 0) && (i < (n - 1))
&& (sieve[arr[i + 1]] == 0))
{
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {6, 3, 4, 5, 6};
int[] sieve = new int[MAX];
int n = arr.length;
int k = 2;
SieveOfEratosthenes(sieve, MAX);
if (isReachable(arr, n, sieve, k))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Python3 program to check if it is
# possible to reach the end of the
# array
import math
# Function for Sieve of Eratosthenes
def SieveOfEratosthenes(sieve, max):
for i in range(0, max):
sieve[i] = 1
sqt = int(math.sqrt(max))
for i in range(2, sqt):
if (sieve[i] == 1):
for j in range(i * 2, max, i):
sieve[j] = 0
# Function to check if it is possible to
# reach end of the array
def isReachable(arr, n, sieve, k):
# store initial value of K
x = k
for i in range(0, n):
if (sieve[arr[i]] != 0):
# Refill K to initial value
k = x
else:
# Decrement k by 1
k -= 1
if ((k <= 0) and (i < (n - 1)) and
(sieve[arr[i + 1]] == 0)):
return 0
return 1
# Driver Code
arr = [ 6, 3, 4, 5, 6 ]
sieve = [0 for x in range(1000000)]
n = len(arr)
k = 2
SieveOfEratosthenes(sieve, 1000000)
ch = isReachable(arr, n, sieve, k)
if (ch):
print("Yes")
else:
print("No")
# This code is contributed by Stream_Cipher
C＃
// C# program to check if it is possible
// to reach the end of the array
using System;
class GFG
{
static int MAX = 1000000;
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes(int []sieve, int max)
{
for (int i = 0; i < max; i++)
{
sieve[i] = 1;
}
for (int i = 2; i * i < max; i++)
{
if (sieve[i] == 1)
{
for (int j = i * 2; j < max; j += i)
{
sieve[j] = 0;
}
}
}
}
// Function to check if it is possible to
// reach end of the array
static bool isReachable(int []arr, int n,
int []sieve, int k)
{
// store initial value of K
int x = k;
for (int i = 0; i < n; i++)
{
if (sieve[arr[i]] == 1)
{
// Refill K to initial value
k = x;
}
else
{
// Decrement k by 1
k -= 1;
}
if ((k <= 0) && (i < (n - 1))
&& (sieve[arr[i + 1]] == 0))
{
return false;
}
}
return true;
}
// Driver Code
static public void Main ()
{
int []arr = {6, 3, 4, 5, 6};
int[] sieve = new int[MAX];
int n = arr.Length;
int k = 2;
SieveOfEratosthenes(sieve, MAX);
if (isReachable(arr, n, sieve, k))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
/* This code contributed by ajit. */
Javascript
输出：
Yes时间复杂度： O(? Max * loglog(Max)) + O(n)