给定两个整数L和R ，任务是计算给定范围内存在的全质数的数量。
A number is said to be Full prime if the number itself is prime and all its digits are also prime.
Examples:
- 53 is Full Prime because it is prime and all its digits (5 and 3) are also prime.
- 13 is not Full Prime because it has a non-prime digit ( 1 is not prime).
例子：
Input: L = 1, R = 100
Output : 8
Explanations: 2 3 5 7 23 37 53 73 are the Full Prime numbers between 1 and 100. Therefore, the count is 8.
Input: L = 200, R = 300
Output: 5
Explanation: 223 227 233 257 277 are the Full Prime numbers between 200 and 300. Therefore, the count is 5.
方法：请按照以下步骤解决问题：
- 只需遍历从L到R的范围。
- 对于范围内的每个数字i ，请检查是否可以将其除以[2，sqrt(i)]范围内的任何数字。如果发现是真的，那么它不是素数。继续下一个数字。
- 否则，请检查其所有数字是否都是质数。如果发现是真的，请增加count 。
- 最后，在遍历范围之后，打印count的值。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to check if a
// number is prime or not
bool isPrime(int num)
{
if (num <= 1)
return false;
for (int i = 2; i * i <= num; i++)
// If a divisor of n exists
if (num % i == 0)
return false;
return true;
}
// Function to check if a
// number is Full Prime or not
bool isFulPrime(int n)
{
// If n is not a prime
if (!isPrime(n))
return false;
// Otherwise
else {
while (n > 0) {
// Extract digit
int rem = n % 10;
// If any digit of n
// is non-prime
if (!(rem == 2 || rem == 3
|| rem == 5 || rem == 7))
return false;
n = n / 10;
}
}
return true;
}
// Function to print count of
// Full Primes in a range [L, R]
int countFulPrime(int L, int R)
{
// Stores count of full primes
int cnt = 0;
for (int i = L; i <= R; i++) {
// Check if i is full prime
if ((i % 2) != 0 && isFulPrime(i)) {
cnt++;
}
}
return cnt;
}
// Driver Code
int main()
{
int L = 1, R = 100;
// Stores count of full primes
int ans = 0;
if (L < 3)
ans++;
cout << ans + countFulPrime(L, R);
return 0;
} Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to check if a
// number is prime or not
static boolean isPrime(int num)
{
if (num <= 1)
return false;
for(int i = 2; i * i <= num; i++)
// If a divisor of n exists
if (num % i == 0)
return false;
return true;
}
// Function to check if a
// number is Full Prime or not
static boolean isFulPrime(int n)
{
// If n is not a prime
if (!isPrime(n))
return false;
// Otherwise
else
{
while (n > 0)
{
// Extract digit
int rem = n % 10;
// If any digit of n
// is non-prime
if (!(rem == 2 || rem == 3 ||
rem == 5 || rem == 7))
return false;
n = n / 10;
}
}
return true;
}
// Function to print count of
// Full Primes in a range [L, R]
static int countFulPrime(int L, int R)
{
// Stores count of full primes
int cnt = 0;
for(int i = L; i <= R; i++)
{
// Check if i is full prime
if ((i % 2) != 0 && isFulPrime(i))
{
cnt++;
}
}
return cnt;
}
// Driver Code
public static void main (String[] args)
{
int L = 1, R = 100;
// Stores count of full primes
int ans = 0;
if (L < 3)
ans++;
System.out.println(ans + countFulPrime(L, R));
}
}
// This code is contributed by sanjoy_62Python3
# Python3 program to implement
# the above approach
# Function to check if a
# number is prime or not
def isPrime(num):
if (num <= 1):
return False
for i in range(2, num + 1):
if i * i > num:
break
# If a divisor of n exists
if (num % i == 0):
return False
return True
# Function to check if a
# number is Full Prime or not
def isFulPrime(n):
# If n is not a prime
if (not isPrime(n)):
return False
# Otherwise
else:
while (n > 0):
# Extract digit
rem = n % 10
# If any digit of n
# is non-prime
if (not (rem == 2 or rem == 3 or
rem == 5 or rem == 7)):
return False
n = n // 10
return True
# Function to prcount of
# Full Primes in a range [L, R]
def countFulPrime(L, R):
# Stores count of full primes
cnt = 0
for i in range(L, R + 1):
# Check if i is full prime
if ((i % 2) != 0 and isFulPrime(i)):
cnt += 1
return cnt
# Driver Code
if __name__ == '__main__':
L = 1
R = 100
# Stores count of full primes
ans = 0
if (L < 3):
ans += 1
print(ans + countFulPrime(L, R))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check if a
// number is prime or not
static bool isPrime(int num)
{
if (num <= 1)
return false;
for(int i = 2; i * i <= num; i++)
// If a divisor of n exists
if (num % i == 0)
return false;
return true;
}
// Function to check if a
// number is Full Prime or not
static bool isFulPrime(int n)
{
// If n is not a prime
if (!isPrime(n))
return false;
// Otherwise
else
{
while (n > 0)
{
// Extract digit
int rem = n % 10;
// If any digit of n
// is non-prime
if (!(rem == 2 || rem == 3 ||
rem == 5 || rem == 7))
return false;
n = n / 10;
}
}
return true;
}
// Function to print count of
// Full Primes in a range [L, R]
static int countFulPrime(int L, int R)
{
// Stores count of full primes
int cnt = 0;
for(int i = L; i <= R; i++)
{
// Check if i is full prime
if ((i % 2) != 0 && isFulPrime(i))
{
cnt++;
}
}
return cnt;
}
// Driver Code
public static void Main (String[] args)
{
int L = 1, R = 100;
// Stores count of full primes
int ans = 0;
if (L < 3)
ans++;
Console.WriteLine(ans + countFulPrime(L, R));
}
}
// This code is contributed by math_lover输出：
8
时间复杂度： O(N 3/2 )
辅助空间： O(1)