给定一个数字，找到既是斐波那契数又是素数的数字(小于或等于n)。
例子：
Input : n = 40
Output: 2 3 5 13
Explanation :
Here, range(upper limit) = 40
Fibonacci series upto n is, 1,
1, 2, 3, 5, 8, 13, 21, 34.
Prime numbers in above series = 2, 3, 5, 13.
Input : n = 100
Output: 2 3 5 13 89
Explanation :
Here, range(upper limit) = 40
Fibonacci series upto n are 1, 1, 2,
3, 5, 8, 13, 21, 34, 55, 89.
Prime numbers in Fibonacci upto n : 2, 3,
5, 13, 89.
一个简单的解决方案是迭代生成所有小于或等于n的斐波那契数。对于每个斐波那契数，请检查其是否为质数。如果是素色，请打印出来。
一个有效的解决方案是使用Sieve生成最多n个素数。生成素数后，如果数字为5i 2 + 4或5i 2 – 4形式，我们可以使用属性为Fibonacci的属性来快速检查素数是否为Fibonacci。 。
以下是上述步骤的执行
C++
// CPP program to print prime numbers present
// in Fibonacci series.
#include
using namespace std;
// Function to check perfect square
bool isSquare(int n)
{
int sr = sqrt(n);
return (sr * sr == n);
}
// Prints all numbers less than or equal to n that
// are both Prime and Fibonacci.
void printPrimeAndFib(int n)
{
// Using Sieve to generate all primes
// less than or equal to n.
bool prime[n + 1];
memset(prime, true, sizeof(prime));
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
// Now traverse through the range and print numbers
// that are both prime and Fibonacci.
for (int i=2; i<=n; i++)
if (prime[i] && (isSquare(5 * i * i + 4) > 0 ||
isSquare(5 * i * i - 4) > 0))
cout << i << " ";
}
// Driver function
int main()
{
int n = 30;
printPrimeAndFib(n);
return 0;
} Java
// Java program to print prime numbers
// present in Fibonacci series.
class PrimeAndFib
{
// Function to check perfect square
Boolean isSquare(int n)
{
int sr = (int)Math.sqrt(n);
return (sr * sr == n);
}
// Prints all numbers less than or equal
// to n that are both Prime and Fibonacci.
static void printPrimeAndFib(int n)
{
// Using Sieve to generate all
// primes less than or equal to n.
Boolean[] prime = new Boolean[n + 1];
// memset(prime, true, sizeof(prime));
for (int p = 0; p <= n; p++)
prime[p] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
// Now traverse through the range and
// print numbers that are both prime
// and Fibonacci.
for (int i=2; i<=n; i++) {
double sqrt = Math.sqrt(5 * i * i + 4);
double sqrt1 = Math.sqrt(5 * i * i - 4);
int x = (int) sqrt;
int y = (int) sqrt1;
if (prime[i]==true && (Math.pow(sqrt,2) ==
Math.pow(x,2) || Math.pow(sqrt1,2) ==
Math.pow(y,2)))
System.out.print(i+" ");
}
}
// driver program
public static void main(String s[])
{
int n = 30;
printPrimeAndFib(n);
}
}
// This code is contributed by Prerna SainiPython 3
# Python 3 program to print
# prime numbers present in
# Fibonacci series.
import math
# Function to check perfect square
def isSquare(n) :
sr = (int)(math.sqrt(n))
return (sr * sr == n)
# Prints all numbers less than
# or equal to n that are
# both Prime and Fibonacci.
def printPrimeAndFib(n) :
# Using Sieve to generate all
# primes less than or equal to n.
prime =[True] * (n + 1)
p = 2
while(p * p <= n ):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True) :
# Update all multiples of p
for i in range(p * 2, n + 1, p) :
prime[i] = False
p = p + 1
# Now traverse through the range
# and print numbers that are
# both prime and Fibonacci.
for i in range(2, n + 1) :
if (prime[i] and (isSquare(5 * i * i + 4) > 0 or
isSquare(5 * i * i - 4) > 0)) :
print(i , " ",end="")
# Driver function
n = 30
printPrimeAndFib(n);
# This code is contributed
# by Nikita Tiwari.C#
// C# program to print prime numbers
// present in Fibonacci series.
using System;
class GFG {
// Function to check perfect square
static bool isSquare(int n)
{
int sr = (int)Math.Sqrt(n);
return (sr * sr == n);
}
// Prints all numbers less than or equal
// to n that are both Prime and Fibonacci.
static void printPrimeAndFib(int n)
{
// Using Sieve to generate all
// primes less than or equal to n.
bool[] prime = new bool[n + 1];
// memset(prime, true, sizeof(prime));
for (int p = 0; p <= n; p++)
prime[p] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
// Now traverse through the range and
// print numbers that are both prime
// and Fibonacci.
for (int i = 2; i <= n; i++) {
double sqrt = Math.Sqrt(5 * i * i + 4);
double sqrt1 = Math.Sqrt(5 * i * i - 4);
int x = (int) sqrt;
int y = (int) sqrt1;
if (prime[i] == true && (Math.Pow(sqrt, 2) ==
Math.Pow(x, 2) || Math.Pow(sqrt1, 2) ==
Math.Pow(y, 2)))
Console.Write(i + " ");
}
}
// driver program
public static void Main()
{
int n = 30;
printPrimeAndFib(n);
}
}
// This code is contributed by Anant Agarwal.PHP
0 ||
isSquare(5 * $i * $i - 4) > 0))
echo $i . " ";
}
// Driver Code
$n = 30;
printPrimeAndFib($n);
// This code is contributed by mits
?>输出：
2 3 5 13