给定一个整数N ，任务是找到N以下所有可截断的素数之和。可截断的质数是一个可左截断的质数(如果连续删除前导(“左”)数字，则所有结果数均为质数)以及可右截断的质数(如果最后一个(“右”)数为依次删除，则所有生成的数字均为质数)。
例如，3797是左截短的素数，因为797、97和7是素数。而且，3797也是右截数素数，因为379、37和3是素数。因此3797是可截短的素数。
例子：
Input: N = 25
Output: 40
2, 3, 5, 7 and 23 are the only truncatable primes below 25.
2 + 3 + 5 + 7 + 23 = 40
Input: N = 40
Output: 77
方法：一种有效的方法是使用Eratosthenes筛子查找所有质数，对于N以下的每个数字，检查其是否为可截断质数。如果是，则将累加到运行总和中。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define N 1000005
// To check if a number is prime or not
bool prime[N];
// Sieve of Eratosthenes
// function to find all prime numbers
void sieve()
{
memset(prime, true, sizeof prime);
prime[1] = false;
prime[0] = false;
for (int i = 2; i < N; i++)
if (prime[i])
for (int j = i * 2; j < N; j += i)
prime[j] = false;
}
// Function to return the sum of
// all truncatable primes below n
int sumTruncatablePrimes(int n)
{
// To store the required sum
int sum = 0;
// Check every number below n
for (int i = 2; i < n; i++) {
int num = i;
bool flag = true;
// Check from right to left
while (num) {
// If number is not prime at any stage
if (!prime[num]) {
flag = false;
break;
}
num /= 10;
}
num = i;
int power = 10;
// Check from left to right
while (num / power) {
// If number is not prime at any stage
if (!prime[num % power]) {
flag = false;
break;
}
power *= 10;
}
// If flag is still true
if (flag)
sum += i;
}
// Return the required sum
return sum;
}
// Driver code
int main()
{
int n = 25;
sieve();
cout << sumTruncatablePrimes(n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static final int N = 1000005;
// To check if a number is prime or not
static boolean prime[] = new boolean[N];
// Sieve of Eratosthenes
// function to find all prime numbers
static void sieve()
{
Arrays.fill(prime, true);
prime[1] = false;
prime[0] = false;
for (int i = 2; i < N; i++)
{
if (prime[i])
{
for (int j = i * 2; j < N; j += i)
{
prime[j] = false;
}
}
}
}
// Function to return the sum of
// all truncatable primes below n
static int sumTruncatablePrimes(int n)
{
// To store the required sum
int sum = 0;
// Check every number below n
for (int i = 2; i < n; i++)
{
int num = i;
boolean flag = true;
// Check from right to left
while (num > 0)
{
// If number is not prime at any stage
if (!prime[num])
{
flag = false;
break;
}
num /= 10;
}
num = i;
int power = 10;
// Check from left to right
while (num / power > 0)
{
// If number is not prime at any stage
if (!prime[num % power])
{
flag = false;
break;
}
power *= 10;
}
// If flag is still true
if (flag)
{
sum += i;
}
}
// Return the required sum
return sum;
}
// Driver code
public static void main(String[] args)
{
int n = 25;
sieve();
System.out.println(sumTruncatablePrimes(n));
}
}
// This code contributed by Rajput-JiPython3
# Python3 implementation of the
# above approach
N = 1000005
# To check if a number is prime or not
prime = [True for i in range(N)]
# Sieve of Eratosthenes
# function to find all prime numbers
def sieve():
prime[1] = False
prime[0] = False
for i in range(2, N):
if (prime[i]==True):
for j in range(i * 2, N, i):
prime[j] = False
# Function to return the sum of
# all truncatable primes below n
def sumTruncatablePrimes(n):
# To store the required sum
sum = 0
# Check every number below n
for i in range(2, n):
num = i
flag = True
# Check from right to left
while (num):
# If number is not prime at any stage
if (prime[num] == False):
flag = False
break
num //= 10
num = i
power = 10
# Check from left to right
while (num // power):
# If number is not prime at any stage
if (prime[num % power] == False):
flag = False
break
power *= 10
# If flag is still true
if (flag==True):
sum += i
# Return the required sum
return sum
# Driver code
n = 25
sieve()
print(sumTruncatablePrimes(n))
# This code is contributed by mohit kumarC#
// C# implementation of the above approach.
using System;
using System.Collections.Generic;
class GFG
{
static int N = 1000005;
// To check if a number is prime or not
static Boolean []prime = new Boolean[N];
// Sieve of Eratosthenes
// function to find all prime numbers
static void sieve()
{
Array.Fill(prime, true);
prime[1] = false;
prime[0] = false;
for (int i = 2; i < N; i++)
{
if (prime[i])
{
for (int j = i * 2; j < N; j += i)
{
prime[j] = false;
}
}
}
}
// Function to return the sum of
// all truncatable primes below n
static int sumTruncatablePrimes(int n)
{
// To store the required sum
int sum = 0;
// Check every number below n
for (int i = 2; i < n; i++)
{
int num = i;
Boolean flag = true;
// Check from right to left
while (num > 0)
{
// If number is not prime at any stage
if (!prime[num])
{
flag = false;
break;
}
num /= 10;
}
num = i;
int power = 10;
// Check from left to right
while (num / power > 0)
{
// If number is not prime at any stage
if (!prime[num % power])
{
flag = false;
break;
}
power *= 10;
}
// If flag is still true
if (flag)
{
sum += i;
}
}
// Return the required sum
return sum;
}
// Driver code
public static void Main(String []args)
{
int n = 25;
sieve();
Console.WriteLine(sumTruncatablePrimes(n));
}
}
// This code has been contributed by Arnab KunduPHP
输出：
40