給定 n，找出有多少個 n 位數字，使得該數字在素數索引處具有素數並且可以被 m 整除？

Question

特殊數字是這樣的數字在素數索引上具有素數（2、3、5、7），在非素數索引上具有非素數。 （例如，15743 - 素數索引 (2, 3, 5) 具有素數 (5, 7, 3)）。 有多少個 n 位數的特殊數也能被 m 整除。

例如，對於 n=2 和 m=2，答案將是 [12,42,62,82,92]，所以 5。

我寫了一個回溯算法，它找到所有特殊數字的排列，然后檢查這些特殊數字中的每一個是否可以被 m 整除並返回計數。 這適用於 n 和 m 的小值，但問題的 n、m 值在 0-500 的范圍內。

 var primes = [2, 3, 5, 7] var nonPrimes = [1, 4, 6, 8, 9] n = 2 // number of digits m = 2 //divisor k = 0 //remainder function rec(index, temp, count) { if (temp.length >= n) { if (Number(temp) % m === k) { /* console.log(temp) */ count += 1 } return count } if (primes.includes(index)) { for (num1 in primes) { temp += primes[num1]; count = rec(index + 1, temp, count) temp = temp.slice(0, -1) } } else if (nonPrimes.includes(index)) { for (num2 in nonPrimes) { temp += nonPrimes[num2]; count = rec(index + 1, temp, count) temp = temp.slice(0, -1) } } return count } console.log("number of n-digit special numbers which are divisible by m with remainder k is ", rec(1, "", 0))

Answer 1

由於逐位遞歸可以使它們相互依賴，因此對小於m的所有余數求解並返回余數 0 的解。給定一個計數表，其中包含“特殊”數字的計數表，當除以m時剩余r到第i個索引。 然后將索引i + 1的行制成表格：

(1) Transform the current row of remainders,
each storing a count, multiplying by 10:

  for remainder r in row:
    new_r = (10 mod m * r) mod m
    new_row[new_r] += row[r]
  
  row = new_row
    
(2) Create new counts by using the new
possible digits:

  initialise new_row with zeros

  for d in allowed digits for this ith row:
    for r in row:
      new_r = (r + d) mod m
      new_row[new_r] += row[r]

例如， n = 2, m = 2 ：

row = [None, None]
# We are aware of the allowed digits
# at the ith row
row[0] = 3  # digits 4, 6 and 8
row[1] = 2  # digits 1, 9
row = [3, 2]

(1) 變換：

  new_row = [0, 0]

  remainder 0:
    new_r = (10 mod 2 * 0) mod 2 = 0
    new_row[0] += row[0] = 3
    
  remainder 1:
    new_r = (10 mod 2 * 1) mod 2 = 0
    new_row[0] += row[1] = 5
  
  row = new_row = [5, 0]

(2) 創建新計數：

  new_row = [0, 0]
  
  d = 2:
    rd = 2 mod 2 = 0
    
    r = 0:
      new_r = (0 + 0) mod 2 = 0
      new_row[0] += row[0] = 5
    r = 1:
      new_r = (1 + 0) mod 2 = 1
      new_row[1] += row[1] = 0
      
  d = 3:  # Similarly for 5, 7
    rd = 3 mod 2 = 1
    
    r = 0:
      new_r = (0 + 1) mod 2 = 1
      new_row[1] += row[0] = 5
    r = 1:
      new_r = (1 + 1) mod 2 = 0
      new_row[0] += row[1] = 5  # unchanged
      
  row = new_row = [5, 15]
  
[12,42,62,82,92]

[13,15,17,
 43,45,47,
 63,65,67,
 83,85,87,
 93,95,97]