计算给定范围内分别在素数和非素数位置具有素数和非素数的数字

给定两个整数L和R ，任务是找出[L, R]范围内在素数位置具有素数而在非素数位置具有非素数的数字的计数。

例子：

Input: L = 5, R = 22  
Output: 7
Explanation: The numbers 6, 8, 9, 12, 13, 15, and 17 have prime digits at prime positions and non-prime digits at non-prime positions.

Input: L = 20, R = 29
Output: 0
Explanation: There are no numbers which have prime digits at prime positions and non-prime digits at non-prime positions.

朴素方法：解决问题的最简单方法是迭代范围[L, R] 。对于每个^第i^个数字，检查数字的数字在素数位置是否为素数，在非素数位置是否为非素数。如果发现为真，则增加计数。最后，打印获得的计数。

时间复杂度： O(R – L + 1) * sqrt(R) * log ₁₀ (R)
辅助空间： O(1)

高效的方法：为了优化上述方法，想法是使用Digit DP。以下是动态规划状态之间的递推关系：

$CntNum(pos, st, tight, prime)$
$= \sum^{9}_{i=0} cntNum(pos + 1, st | (i > 0), tight \& (i==end), prime + x)$
If i is a prime at prime digits or non-prime at non-prime digits, then x = 1 
 
pos: Stores position of digits 
prime: Check if prime digits are present at prime positions and non-prime digits at non-prime positions are present or not. 
st: check if a number contains any leading 0. 
end: Maximum possible digits at current position

 

请按照以下步骤解决问题：

• 初始化一个 4D 数组，比如dp[pos][st][tight][prime] 。
• 使用记忆计算数字R的dp[pos][st][tight][prime]值，比如cntR 。
• 使用记忆计算数字L – 1的dp[pos][st][tight][prime]值，比如cntL 。
• 最后，打印(cntR – cntL)的值。

下面是上述方法的实现：

C++14

// C++ program for the above approach
 
#include 
using namespace std;
 
// Store digits of a number
vector num;
 
// Store overlapping subproblems
long long int dp[19][2][2][19];
 
// Function to check if a
// number is prime or not
bool isPrime(long long int n)
{
 
    // If n is less than
    // or equal to 1
    if (n <= 1)
        return false;
 
    // If n is less than
    // or equal to 3
    if (n <= 3)
        return true;
 
    // If n is a multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    // Iterate over the range [5, n]
    for (long long int i = 5; i * i <= n;
         i = i + 6) {
 
        // If n is a multiple of i or (i + 2)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    }
 
    return true;
}
 
// Function to count the required
// numbers from the given range
long long cntNum(long long pos, long long st,
                 long long tight, long long prime)
{
 
    // Base Case
    if (pos == num.size())
        return 1;
 
    // If the subproblems already computed
    if (dp[pos][st][tight][prime] != -1)
        return dp[pos][st][tight][prime];
 
    long long int res = 0;
 
    // Stores maximum possible
    // at current digits
    long long end = (tight == 0) ? num[pos] : 9;
 
    // Iterate over all possible digits
    // at current position
    for (long long i = 0; i <= end; i++) {
 
        // Check if i is the maximum possible
        // digit at current position or not
        long long ntight = (i < end) ? 1 : tight;
 
        // Check if a number contains
        // leading 0s or not
        long long int nzero = (i != 0) ? 1 : st;
 
        // If number has only leading zeros
        // and digit is non-zero
        if ((nzero == 1) && isPrime(i) && isPrime(prime)) {
 
            // Prime digits at prime positions
            res += cntNum(pos + 1, nzero,
                          ntight, prime + 1);
        }
 
        if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) {
 
            // Non-prime digits at
            // non-prime positions
            res += cntNum(pos + 1, nzero,
                          ntight, prime + 1);
        }
 
        // If the number has only leading zeros
        // and i is zero,
        if (nzero == 0)
            res += cntNum(pos + 1, nzero,
                          ntight, prime);
    }
    return dp[pos][st][tight][prime] = res;
}
 
// Function to find count of numbers in
// range [0, b] whose digits are prime
// at prime and non-prime at non-prime pos
long long int cntZeroRange(long long int b)
{
 
    num.clear();
 
    // Insert digits of a number, b
    while (b > 0) {
        num.push_back(b % 10);
        b /= 10;
    }
 
    // Reversing the digits in num
    reverse(num.begin(), num.end());
 
    // Initializing dp with -1
    memset(dp, -1, sizeof(dp));
 
    long long int res = cntNum(0, 0, 0, 1);
 
    // Returning the value
    return res;
}
 
// Driver Code
int main()
{
 
    // Given range, [L, R]
    long long int L = 5, R = 22;
 
    // Function Call
    long long int res
        = cntZeroRange(R) - cntZeroRange(L - 1);
 
    // Print answer
    cout << res << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Store digits of a number
static Vector num = new Vector<>();
 
// Store overlapping subproblems
static int [][][][]dp = new int[19][2][2][19];
 
// Function to check if a
// number is prime or not
static boolean isPrime(int n)
{
 
    // If n is less than
    // or equal to 1
    if (n <= 1)
        return false;
 
    // If n is less than
    // or equal to 3
    if (n <= 3)
        return true;
 
    // If n is a multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    // Iterate over the range [5, n]
    for (int i = 5; i * i <= n;
         i = i + 6) {
 
        // If n is a multiple of i or (i + 2)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    }
    return true;
}
 
// Function to count the required
// numbers from the given range
static int cntNum(int pos, int st,
                 int tight, int prime)
{
 
    // Base Case
    if (pos == num.size())
        return 1;
 
    // If the subproblems already computed
    if (dp[pos][st][tight][prime] != -1)
        return dp[pos][st][tight][prime];
    int res = 0;
 
    // Stores maximum possible
    // at current digits
    int end = (tight == 0) ? num.get(pos) : 9;
 
    // Iterate over all possible digits
    // at current position
    for (int i = 0; i <= end; i++)
    {
 
        // Check if i is the maximum possible
        // digit at current position or not
        int ntight = (i < end) ? 1 : tight;
 
        // Check if a number contains
        // leading 0s or not
        int nzero = (i != 0) ? 1 : st;
 
        // If number has only leading zeros
        // and digit is non-zero
        if ((nzero == 1) && isPrime(i) && isPrime(prime)) {
 
            // Prime digits at prime positions
            res += cntNum(pos + 1, nzero,
                          ntight, prime + 1);
        }
 
        if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) {
 
            // Non-prime digits at
            // non-prime positions
            res += cntNum(pos + 1, nzero,
                          ntight, prime + 1);
        }
 
        // If the number has only leading zeros
        // and i is zero,
        if (nzero == 0)
            res += cntNum(pos + 1, nzero,
                          ntight, prime);
    }
    return dp[pos][st][tight][prime] = res;
}
 
// Function to find count of numbers in
// range [0, b] whose digits are prime
// at prime and non-prime at non-prime pos
static int cntZeroRange(int b)
{
 
    num.clear();
 
    // Insert digits of a number, b
    while (b > 0) {
        num.add(b % 10);
        b /= 10;
    }
 
    // Reversing the digits in num
    Collections.reverse(num);
 
    // Initializing dp with -1
    for (int i = 0; i < 19; i++)
         for (int j = 0; j < 2; j++)
             for (int k = 0; k < 2; k++)
                 for (int l = 0; l < 19; l++)
                     dp[i][j][k][l] = -1;
 
    int res = cntNum(0, 0, 0, 1);
 
    // Returning the value
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given range, [L, R]
    int L = 5, R = 22;
 
    // Function Call
    int res
        = cntZeroRange(R) - cntZeroRange(L - 1);
 
    // Print answer
    System.out.print(res +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
from math import ceil, sqrt
 
# Function to check if a
# number is prime or not
def isPrime(n):
     
    # If n is less than
    # or equal to 1
    if (n <= 1):
        return False
 
    # If n is less than
    # or equal to 3
    if (n <= 3):
        return True
 
    # If n is a multiple of 2 or 3
    if (n % 2 == 0 or n % 3 == 0):
        return False
 
    # Iterate over the range [5, n]
    for i in range(5, ceil(sqrt(n)), 6):
         
        # If n is a multiple of i or (i + 2)
        if (n % i == 0 or n % (i + 2) == 0):
            return False
 
    return True
 
# Function to count the required
# numbers from the given range
def cntNum(pos, st, tight, prime):
     
    global dp, num
     
    if (pos == len(num)):
        return 1
 
    # If the subproblems already computed
    if (dp[pos][st][tight][prime] != -1):
        return dp[pos][st][tight][prime]
 
    res = 0
 
    # Stores maximum possible
    # at current digits
    end = num[pos] if (tight == 0) else  9
 
    # Iterate over all possible digits
    # at current position
    for i in range(end + 1):
         
        # Check if i is the maximum possible
        # digit at current position or not
        ntight = 1 if (i < end) else tight
 
        # Check if a number contains
        # leading 0s or not
        nzero = 1 if (i != 0) else st
 
        # If number has only leading zeros
        # and digit is non-zero
        if ((nzero == 1) and isPrime(i) and
                             isPrime(prime)):
                                  
            # Prime digits at prime positions
            res += cntNum(pos + 1, nzero, ntight,
                        prime + 1)
 
        if ((nzero == 1) and isPrime(i) == False and
                         isPrime(prime) == False):
 
            # Non-prime digits at
            # non-prime positions
            res += cntNum(pos + 1, nzero, ntight,
                        prime + 1)
 
        # If the number has only leading zeros
        # and i is zero,
        if (nzero == 0):
            res += cntNum(pos + 1, nzero,
                          ntight, prime)
                           
    dp[pos][st][tight][prime] = res
     
    return dp[pos][st][tight][prime]
 
# Function to find count of numbers in
# range [0, b] whose digits are prime
# at prime and non-prime at non-prime pos
def cntZeroRange(b):
     
    global num, dp
 
    num.clear()
     
    while (b > 0):
        num.append(b % 10)
        b //= 10
 
    # Reversing the digits in num
    num = num[::-1]
 
    # print(num)
    dp = [[[[-1 for i in range(19)]
                for i in range(2)]
                for i in range(2)]
                for i in range(19)]
 
    res = cntNum(0, 0, 0, 1)
 
    # Returning the value
    return res
 
# Driver Code
if __name__ == '__main__':
 
    dp = [[[[-1 for i in range(19)]
                for i in range(2)]
                for i in range(2)]
                for i in range(19)]
    L, R, num = 5, 22, []
 
    # Function Call
    res = cntZeroRange(R) - cntZeroRange(L - 1)
 
    # Print answer
    print(res)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Store digits of a number
  static List num = new List();
 
  // Store overlapping subproblems
  static int[, , , ] dp = new int[19, 2, 2, 19];
 
  // Function to check if a
  // number is prime or not
  static bool isPrime(int n)
  {
 
    // If n is less than
    // or equal to 1
    if (n <= 1)
      return false;
 
    // If n is less than
    // or equal to 3
    if (n <= 3)
      return true;
 
    // If n is a multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
      return false;
 
    // Iterate over the range [5, n]
    for (int i = 5; i * i <= n; i = i + 6) {
 
      // If n is a multiple of i or (i + 2)
      if (n % i == 0 || n % (i + 2) == 0)
        return false;
    }
    return true;
  }
 
  // Function to count the required
  // numbers from the given range
  static int cntNum(int pos, int st, int tight, int prime)
  {
 
    // Base Case
    if (pos == num.Count)
      return 1;
 
    // If the subproblems already computed
    if (dp[pos, st, tight, prime] != -1)
      return dp[pos, st, tight, prime];
    int res = 0;
 
    // Stores maximum possible
    // at current digits
    int end = (tight == 0) ? num[pos] : 9;
 
    // Iterate over all possible digits
    // at current position
    for (int i = 0; i <= end; i++) {
 
      // Check if i is the maximum possible
      // digit at current position or not
      int ntight = (i < end) ? 1 : tight;
 
      // Check if a number contains
      // leading 0s or not
      int nzero = (i != 0) ? 1 : st;
 
      // If number has only leading zeros
      // and digit is non-zero
      if ((nzero == 1) && isPrime(i)
          && isPrime(prime)) {
 
        // Prime digits at prime positions
        res += cntNum(pos + 1, nzero, ntight,
                      prime + 1);
      }
 
      if ((nzero == 1) && !isPrime(i)
          && !isPrime(prime)) {
 
        // Non-prime digits at
        // non-prime positions
        res += cntNum(pos + 1, nzero, ntight,
                      prime + 1);
      }
 
      // If the number has only leading zeros
      // and i is zero,
      if (nzero == 0)
        res += cntNum(pos + 1, nzero, ntight,
                      prime);
    }
    return dp[pos, st, tight, prime] = res;
  }
 
  // Function to find count of numbers in
  // range [0, b] whose digits are prime
  // at prime and non-prime at non-prime pos
  static int cntZeroRange(int b)
  {
 
    num.Clear();
 
    // Insert digits of a number, b
    while (b > 0) {
      num.Add(b % 10);
      b /= 10;
    }
 
    // Reversing the digits in num
    num.Reverse();
 
    // Initializing dp with -1
    for (int i = 0; i < 19; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++)
          for (int l = 0; l < 19; l++)
            dp[i, j, k, l] = -1;
 
    int res = cntNum(0, 0, 0, 1);
 
    // Returning the value
    return res;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given range, [L, R]
    int L = 5, R = 22;
 
    // Function Call
    int res = cntZeroRange(R) - cntZeroRange(L - 1);
 
    // Print answer
    Console.WriteLine(res + "\n");
  }
}
 
// This code is contributed by chitranayal.

Javascript

 
// JavaScript program for the above approach
 
// Store digits of a number
let num = [];
 
// Store overlapping subproblems
let dp = new Array(19);
 
 
// Function to check if a
// number is prime or not
function isPrime(n)
{
    // If n is less than
    // or equal to 1
    if (n <= 1)
        return false;
  
    // If n is less than
    // or equal to 3
    if (n <= 3)
        return true;
  
    // If n is a multiple of 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
  
    // Iterate over the range [5, n]
    for (let i = 5; i * i <= n;
         i = i + 6) {
  
        // If n is a multiple of i or (i + 2)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    }
    return true;
}
 
// Function to count the required
// numbers from the given range
function cntNum(pos,st,tight,prime)
{
    // Base Case
    if (pos == num.length)
        return 1;
  
    // If the subproblems already computed
    if (dp[pos][st][tight][prime] != -1)
        return dp[pos][st][tight][prime];
    let res = 0;
  
    // Stores maximum possible
    // at current digits
    let end = (tight == 0) ? num[pos] : 9;
  
    // Iterate over all possible digits
    // at current position
    for (let i = 0; i <= end; i++)
    {
  
        // Check if i is the maximum possible
        // digit at current position or not
        let ntight = (i < end) ? 1 : tight;
  
        // Check if a number contains
        // leading 0s or not
        let nzero = (i != 0) ? 1 : st;
  
        // If number has only leading zeros
        // and digit is non-zero
        if ((nzero == 1) && isPrime(i) && isPrime(prime)) {
  
            // Prime digits at prime positions
            res += cntNum(pos + 1, nzero,
                          ntight, prime + 1);
        }
  
        if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) {
  
            // Non-prime digits at
            // non-prime positions
            res += cntNum(pos + 1, nzero,
                          ntight, prime + 1);
        }
  
        // If the number has only leading zeros
        // and i is zero,
        if (nzero == 0)
            res += cntNum(pos + 1, nzero,
                          ntight, prime);
    }
    return dp[pos][st][tight][prime] = res;
}
 
// Function to find count of numbers in
// range [0, b] whose digits are prime
// at prime and non-prime at non-prime pos
function cntZeroRange(b)
{
    num=[];
  
    // Insert digits of a number, b
    while (b > 0) {
        num.push(b % 10);
        b = Math.floor(b/10);
    }
  
    // Reversing the digits in num
    num.reverse();
     for(let i=0;i<19;i++)
{
    dp[i]=new Array(2);
    for(let j=0;j<2;j++)
    {
        dp[i][j]=new Array(2);
        for(let k=0;k<2;k++)
        {
            dp[i][j][k]=new Array(19);
            for(let l=0;l<19;l++)
            {
                dp[i][j][k][l]=-1;
            }
        }
    }
}
     
     
    let res = cntNum(0, 0, 0, 1);
  
    // Returning the value
    return res;
}
 
// Driver Code
// Given range, [L, R]
let L = 5, R = 22;
 
// Function Call
let res
= cntZeroRange(R) - cntZeroRange(L - 1);
 
// Print answer
document.write(res +"<br>");
 
 
// This code is contributed by avanitrachhadiya2155
 

输出：

7

时间复杂度： O(log ₁₀ (R) * log ₁₀ (L) sqrt(log ₁₀ (R))* 10 * 4))
辅助空间： O(log ₁₀ (R) * log ₁₀ (L) * 4)