给定一个正整数N，检查它是否可以表示为两个半素数之和。
半质数如果一个数字可以表示为两个质数(不一定是互斥的)的乘积，则称该数为半质数。
1 -100范围内的半素是：
4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95.
例子：
Input : N = 30
Output: YES
Explanation: 30 can be expressed as '15 + 15'
15 is semi-primes as 15 is a product of two primes 3 and 5.
Input : N = 45
Output : YES
Explanation: 45 can be expressed as '35 + 10'
35 and 10 are also semi-primes as it can a be expressed
as product of two primes:
35 = 5 * 7
10 = 2 * 5
先决条件：
- 半素数
一个简单的解决方案是遍历i = 1到
并检查i和Ni是否为半素数。如果是，请打印i和ni。
一个有效的解决方案是预先计算数组中的半素数到给定范围，然后遍历该半素数数组，并检查n-arr [i]是否为半素数。因为，如果n-arr [i]也是半素数，则arr [i]已经是一个半素数，则n可以表示为两个半素数的总和。
下面是上述方法的实现：
C++
// CPP Code to check if an integer
// can be expressed as sum of
// two semi-primes
#include
using namespace std;
#define MAX 1000000
vector arr;
bool sprime[MAX];
// Utility function to compute
// semi-primes in a range
void computeSemiPrime()
{
memset(sprime, false, sizeof(sprime));
for (int i = 2; i < MAX; i++) {
int cnt = 0;
int num = i;
for (int j = 2; cnt < 2 && j * j <= num; ++j) {
while (num % j == 0) {
num /= j, ++cnt; // Increment count
// of prime numbers
}
}
// If number is greater than 1, add it to
// the count variable as it indicates the
// number remain is prime number
if (num > 1)
++cnt;
// if count is equal to '2' then
// number is semi-prime
if (cnt == 2) {
sprime[i] = true;
arr.push_back(i);
}
}
}
// Utility function to check
// if a number sum of two
// semi-primes
bool checkSemiPrime(int n)
{
int i = 0;
while (arr[i] <= n / 2) {
// arr[i] is already a semi-prime
// if n-arr[i] is also a semi-prime
// then we a number can be expressed as
// sum of two semi-primes
if (sprime[n - arr[i]]) {
return true;
}
i++;
}
return false;
}
// Driver code
int main()
{
computeSemiPrime();
int n = 30;
if (checkSemiPrime(n))
cout << "YES";
else
cout << "NO";
return 0;
} Java
// Java Code to check if an integer
// can be expressed as sum of
// two semi-primes
import java.util.*;
class GFG {
static final int MAX = 1000000;
static Vector arr = new Vector<>();
static boolean[] sprime = new boolean[MAX];
// Utility function to compute
// semi-primes in a range
static void computeSemiPrime()
{
for (int i = 0; i < MAX; i++)
sprime[i] = false;
for (int i = 2; i < MAX; i++) {
int cnt = 0;
int num = i;
for (int j = 2; cnt < 2 && j * j <= num; ++j) {
while (num % j == 0) {
num /= j;
++cnt;
// Increment count
// of prime numbers
}
}
// If number is greater than 1, add it to
// the count variable as it indicates the
// number remain is prime number
if (num > 1)
++cnt;
// if count is equal to '2' then
// number is semi-prime
if (cnt == 2) {
sprime[i] = true;
arr.add(i);
}
}
}
// Utility function to check
// if a number is sum of two
// semi-primes
static boolean checkSemiPrime(int n)
{
int i = 0;
while (arr.get(i) <= n / 2) {
// arr[i] is already a semi-prime
// if n-arr[i] is also a semi-prime
// then a number can be expressed as
// sum of two semi-primes
if (sprime[n - arr.get(i)]) {
return true;
}
i++;
}
return false;
}
// Driver code
public static void main(String[] args)
{
computeSemiPrime();
int n = 30;
if (checkSemiPrime(n))
System.out.println("YES");
else
System.out.println("NO");
}
} Python3
# Python3 Code to check if an integer can
# be expressed as sum of two semi-primes
MAX = 10000
arr = []
sprime = [False] * (MAX)
# Utility function to compute
# semi-primes in a range
def computeSemiPrime():
for i in range(2, MAX):
cnt, num, j = 0, i, 2
while cnt < 2 and j * j <= num:
while num % j == 0:
num /= j
# Increment count of prime numbers
cnt += 1
j += 1
# If number is greater than 1, add it
# to the count variable as it indicates
# the number remain is prime number
if num > 1:
cnt += 1
# if count is equal to '2' then
# number is semi-prime
if cnt == 2:
sprime[i] = True
arr.append(i)
# Utility function to check
# if a number sum of two
# semi-primes
def checkSemiPrime(n):
i = 0
while arr[i] <= n // 2:
# arr[i] is already a semi-prime
# if n-arr[i] is also a semi-prime
# then a number can be expressed as
# sum of two semi-primes
if sprime[n - arr[i]] == True:
return True
i += 1
return False
# Driver code
if __name__ == "__main__":
computeSemiPrime()
n = 30
if checkSemiPrime(n) == True:
print("YES")
else:
print("NO")
# This code is contributed by
# Rituraj JainC#
// C# Code to check if an integer
// can be expressed as sum of
// two semi-primes
using System.Collections.Generic;
class GFG
{
static int MAX = 1000000;
static List arr = new List();
static bool[] sprime = new bool[MAX];
// Utility function to compute
// semi-primes in a range
static void computeSemiPrime()
{
for (int i = 0; i < MAX; i++)
sprime[i] = false;
for (int i = 2; i < MAX; i++)
{
int cnt = 0;
int num = i;
for (int j = 2; cnt < 2 && j * j <= num; ++j)
{
while (num % j == 0)
{
num /= j;
++cnt;
// Increment count
// of prime numbers
}
}
// If number is greater than 1, add it to
// the count variable as it indicates the
// number remain is prime number
if (num > 1)
++cnt;
// if count is equal to '2' then
// number is semi-prime
if (cnt == 2)
{
sprime[i] = true;
arr.Add(i);
}
}
}
// Utility function to check
// if a number is sum of two
// semi-primes
static bool checkSemiPrime(int n)
{
int i = 0;
while (arr[i] <= n / 2)
{
// arr[i] is already a semi-prime
// if n-arr[i] is also a semi-prime
// then a number can be expressed as
// sum of two semi-primes
if (sprime[n - arr[i]])
{
return true;
}
i++;
}
return false;
}
// Driver code
public static void Main()
{
computeSemiPrime();
int n = 30;
if (checkSemiPrime(n))
System.Console.WriteLine("YES");
else
System.Console.WriteLine("NO");
}
}
// This code is contributed by mits 输出：
YES