Blum Integer是一个半素数，假设p和q是两个因子(即n = p * q)，它们(p和q)的形式为4t + 3，其中t是某个整数。
前几个Blum Integer是21、33、57、69、77、93、129、133、141、161、177，…
注意：由于两个因子都应为半素数，因此偶数不能是Blum整数，也不能是20以下的数字，
因此，无论是否为Blum Integer，我们都只需检查大于20的奇数整数。
例子 ：
Input: 33
Output: Yes
Explanation: 33 = 3 * 11, 3 and 11 are both
semi-primes as well as of the form 4t + 3
for t = 0, 2
Input: 77
Output: Yes
Explanation: 77 = 7 * 11, 7 and 11 both are
semi-prime as well as of the form 4t + 3
for t = 1, 2
Input: 25
Output: No
Explanation: 25 = 5*5, 5 and 5 are both
semi-prime but are not of the form 4t
+ 3, Hence 25 is not a Blum integer.方法：对于给定的大于20的奇整数，我们计算从1到该奇整数的质数。如果我们找到任何将那个奇数整数除以它的商均是素数的质数，并且对于某个整数，其形式为4t + 3，那么给定的奇数整数就是Blum Integer。
下面是上述方法的实现：
C++
// CPP program to check if a number is a Blum
// integer
#include
using namespace std;
// Function to cheek if number is Blum Integer
bool isBlumInteger(int n)
{
bool prime[n + 1];
memset(prime, true, sizeof(prime));
// to store prime numbers from 2 to n
for (int i = 2; i * i <= n; i++) {
// If prime[i] is not
// changed, then it is a prime
if (prime[i] == true) {
// Update all multiples of p
for (int j = i * 2; j <= n; j += i)
prime[j] = false;
}
}
// to check if the given odd integer
// is Blum Integer or not
for (int i = 2; i <= n; i++) {
if (prime[i]) {
// checking the factors
// are of 4t+3 form or not
if ((n % i == 0) && ((i - 3) % 4) == 0) {
int q = n / i;
return (q != i && prime[q] &&
(q - 3) % 4 == 0);
}
}
}
return false;
}
// driver code
int main()
{
// give odd integer greater than 20
int n = 249;
if (isBlumInteger(n))
cout << "Yes";
else
cout << "No";
} Java
// Java implementation to check If
// a number is a Blum integer
import java.util.*;
class GFG {
public static boolean isBlumInteger(int n)
{
boolean prime[] = new boolean[n + 1];
for (int i = 0; i < n; i++)
prime[i] = true;
// to store prime numbers from 2 to n
for (int i = 2; i * i <= n; i++) {
// If prime[i] is not changed,
// then it is a prime
if (prime[i] == true) {
// Update all multiples of p
for (int j = i * 2; j <= n; j += i)
prime[j] = false;
}
}
// to check if the given odd integer
// is Blum Integer or not
for (int i = 2; i <= n; i++) {
if (prime[i]) {
// checking the factors are
// of 4t + 3 form or not
if ((n % i == 0) && ((i - 3) % 4) == 0) {
int q = n / i;
return (q != i && prime[q] &&
(q - 3) % 4 == 0);
}
}
}
return false;
}
// driver code
public static void main(String[] args)
{
// give odd integer greater than 20
int n = 249;
if (isBlumInteger(n) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}Python 3
# python 3 program to check if a
# number is a Blum integer
# Function to cheek if number
# is Blum Integer
def isBlumInteger(n):
prime = [True]*(n + 1)
# to store prime numbers from 2 to n
i = 2
while (i * i <= n):
# If prime[i] is not
# changed, then it is a prime
if (prime[i] == True) :
# Update all multiples of p
for j in range(i * 2, n + 1, i):
prime[j] = False
i = i + 1
# to check if the given odd integer
# is Blum Integer or not
for i in range(2, n + 1) :
if (prime[i]) :
# checking the factors
# are of 4t+3 form or not
if ((n % i == 0) and
((i - 3) % 4) == 0):
q = int(n / i)
return (q != i and prime[q]
and (q - 3) % 4 == 0)
return False
# driver code
# give odd integer greater than 20
n = 249
if (isBlumInteger(n)):
print("Yes")
else:
print("No")
# This code is contributed by Smitha.C#
// C# implementation to check If
// a number is a Blum integer
using System;
class GFG {
public static bool isBlumInteger(int n)
{
bool[] prime = new bool[n + 1];
for (int i = 0; i < n; i++)
prime[i] = true;
// to store prime numbers from 2 to n
for (int i = 2; i * i <= n; i++) {
// If prime[i] is not changed,
// then it is a prime
if (prime[i] == true) {
// Update all multiples of p
for (int j = i * 2; j <= n; j += i)
prime[j] = false;
}
}
// to check if the given odd integer
// is Blum Integer or not
for (int i = 2; i <= n; i++) {
if (prime[i]) {
// checking the factors are
// of 4t + 3 form or not
if ((n % i == 0) && ((i - 3) % 4) == 0)
{
int q = n / i;
return (q != i && prime[q] &&
(q - 3) % 4 == 0);
}
}
}
return false;
}
// Driver code
static public void Main ()
{
// give odd integer greater than 20
int n = 249;
if (isBlumInteger(n) == true)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Ajit.PHP
输出：
Yes