如果开明数以其唯一质数的串联开头，则它是一个复合数N
一些开明的数字是：
250, 256, 2048, 2176, 2304, 2500, 2560…
检查N是否为开明数字
给定数字N ，任务是检查N是否为开明数字。如果N是一个开明数字，则打印“是”，否则打印“否” 。
例子：
Input: N = 2500
Output: Yes
Explanation:
factorization of 25 = 2^2 * 5^4
and N begins also with ’25’.
Input: N = 25
Output: No
方法：想法是检查N是否为合成的，如果不是合成的，则返回false，否则将字符串N中所有不同的素数因子连接起来。还要将数字N转换为字符串。现在我们只需要检查不同素数的级联是否是数字N的前缀，如果是则打印“是”，否则打印“否” 。
下面是上述方法的实现：
C++
// C++ implementation of the
// above approach
#include
#include
using namespace std;
// Function to check if N is a
// Composite Number
bool isComposite(int n)
{
// Corner cases
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for(int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to return concatenation of
// distinct prime factors of a given number n
int concatenatePrimeFactors(int n)
{
char concatenate;
// Handle prime factor 2 explicitly
// so that can optimally handle
// other prime factors.
if (n % 2 == 0)
{
concatenate += char(2);
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i + 2)
for(int i = 3; i <= sqrt(n); i = i + 2)
{
// While i divides n, print
// i and divide n
if (n % i == 0)
{
concatenate += i;
while (n % i == 0)
n = n / i;
}
}
// This condition is to handle the
// case when n is a prime number
// greater than 2
if (n > 2)
concatenate += n;
return concatenate;
}
// Function to check if a number is
// is an enlightened number
bool isEnlightened(int N)
{
// Number should not be prime
if (!isComposite(N))
return false;
// Converting N to string
char num = char(N);
// Function call
char prefixConc = concatenatePrimeFactors(N);
return int(prefixConc);
}
// Driver code
int main()
{
int n = 250;
if (isEnlightened(n))
cout << "Yes";
else
cout << "No";
}
// This code is contributed by adityakumar27200 Java
// Java implementation of the
// above approach
import java.util.*;
class GFG {
// Function to check if N is a
// Composite Number
static boolean isComposite(int n)
{
// Corner cases
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to return concatenation of
// distinct prime factors of a given number n
static String concatenatePrimeFactors(int n)
{
String concatenate = "";
// Handle prime factor 2 explicitly
// so that can optimally handle
// other prime factors.
if (n % 2 == 0) {
concatenate += "2";
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i + 2)
for (int i = 3; i <= Math.sqrt(n); i = i + 2) {
// While i divides n, print
// i and divide n
if (n % i == 0) {
concatenate += i;
while (n % i == 0)
n = n / i;
}
}
// This condition is to handle the
// case when n is a prime number
// greater than 2
if (n > 2)
concatenate += n;
return concatenate;
}
// Function to check if a number is
// is an enlightened number
static boolean isEnlightened(int N)
{
// number should not be prime
if (!isComposite(N))
return false;
// converting N to string
String num = String.valueOf(N);
// function call
String prefixConc
= concatenatePrimeFactors(N);
return num.startsWith(prefixConc);
}
// Driver code
public static void main(String args[])
{
int n = 250;
if (isEnlightened(n))
System.out.println("Yes");
else
System.out.println("No");
}
}Python3
# Python3 implementation of the
# above approach
import math
# Function to check if N is a
# Composite Number
def isComposite(n):
# Corner cases
if n <= 3:
return False
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return True
i = 5
while(i * i <= n):
if(n % i == 0 or n % (i + 2) == 0):
return True
i = i + 6
return False
# Function to return concatenation of
# distinct prime factors of a given number n
def concatenatePrimeFactors(n):
concatenate = ""
# Handle prime factor 2 explicitly
# so that can optimally handle
# other prime factors.
if(n % 2 == 0):
concatenate += "2"
while(n % 2 == 0):
n = int(n / 2)
# n must be odd at this point.
# So we can skip one element
# (Note i = i + 2)
i = 3
while(i <= int(math.sqrt(n))):
# While i divides n, print
# i and divide n
if(n % i == 0):
concatenate += str(i)
while(n % i == 0):
n = int(n / i)
i = i + 2
# This condition is to handle the
# case when n is a prime number
# greater than 2
if(n > 2):
concatenate += str(n)
return concatenate
# Function to check if a number is
# is an enlightened number
def isEnlightened(N):
# Number should not be prime
if(not isComposite(N)):
return False
# Converting N to string
num = str(N)
# Function call
prefixConc = concatenatePrimeFactors(N)
return int(prefixConc)
# Driver code
if __name__=="__main__":
n = 250
if(isEnlightened(n)):
print("Yes")
else:
print("No")
# This code is contributed by adityakumar27200C#
// C# implementation of the
// above approach
using System;
class GFG{
// Function to check if N is a
// Composite Number
static bool isComposite(int n)
{
// Corner cases
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for(int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function to return concatenation of
// distinct prime factors of a given number n
static String concatenatePrimeFactors(int n)
{
String concatenate = "";
// Handle prime factor 2 explicitly
// so that can optimally handle
// other prime factors.
if (n % 2 == 0)
{
concatenate += "2";
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i + 2)
for(int i = 3; i <= Math.Sqrt(n);
i = i + 2)
{
// While i divides n, print
// i and divide n
if (n % i == 0)
{
concatenate += i;
while (n % i == 0)
n = n / i;
}
}
// This condition is to handle the
// case when n is a prime number
// greater than 2
if (n > 2)
concatenate += n;
return concatenate;
}
// Function to check if a number is
// is an enlightened number
static bool isEnlightened(int N)
{
// Number should not be prime
if (!isComposite(N))
return false;
// Converting N to string
String num = String.Join("", N);
// Function call
String prefixConc = concatenatePrimeFactors(N);
return num.StartsWith(prefixConc);
}
// Driver code
public static void Main(String []args)
{
int n = 250;
if (isEnlightened(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Rajput-Ji输出：
Yes
时间复杂度： O(n)
参考：http://www.numbersaplenty.com/set/enlightened_number/