在只有3和4的数字系统中找到第n个数字

给定只有3和4的数字系统。在数字系统中找到第n个数字。号码系统中的前几个数字是：3、4、33、34、43、44、333、334、343、344、433、434、443、444、3333、3334、3343、3344、3433、3434、3443 ，3444，…
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我们可以使用(i-1)位数字生成所有i位数字。这个想法是先在所有带有(i-1)数字的数字前添加一个’3’作为前缀，然后再添加一个’4’。例如，带有2位数字的数字是33、34、43和44。带有3位数字的数字是333、334、343、344、433、434、443和444，可以通过先添加3作为前缀来生成，然后4。
以下是详细步骤。

1) Create an array 'arr[]' of strings size n+1. 
2) Initialize arr[0] as empty string. (Number with 0 digits)
3) Do following while array size is smaller than or equal to n
.....a) Generate numbers by adding a 3 as prefix to the numbers generated 
        in previous iteration.  Add these numbers to arr[]
.....a) Generate numbers by adding a 4 as prefix to the numbers generated 
        in previous iteration. Add these numbers to arr[]

感谢kaushik Lele在这里的评论中建议这个想法。以下是相同的C++实现。

C++

// C++ program to find n'th number
// in a number system with
// only 3 and 4
#include 
using namespace std;
 
// Function to find n'th number
// in a number system with only
// 3 and 4
void find(int n)
{
   
    // An array of strings to
    // store first n numbers. arr[i]
    // stores i'th number
    string arr[n + 1];
   
    // arr[0] stores the empty string (String
    // with 0 digits)
    arr[0] = "";
 
    // size indicates number of
    // current elements in arr[]. m
    // indicates number of elements
    // added to arr[] in
    // previous iteration.
    int size = 1, m = 1;
 
    // Every iteration of following
    // loop generates and adds
    // 2*m numbers to arr[] using 
    // the m numbers generated in
    // previous iteration.
    while (size <= n) {
       
        // Consider all numbers added
        // in previous iteration,
        // add a prefix "3" to them and
        // add new numbers to
        // arr[]
        for (int i = 0; i < m && (size + i) <= n; i++)
            arr[size + i] = "3" + arr[size - m + i];
 
        // Add prefix "4" to numbers
        // of previous iteration
        // and add new numbers to arr[]
        for (int i = 0; i < m && (size + m + i) <= n; i++)
            arr[size + m + i] = "4" + arr[size - m + i];
 
        // Update no. of elements added in previous
        // iteration
        m = m << 1; // Or m = m*2;
 
        // Update size
        size = size + m;
    }
    cout << arr[n] << endl;
}
 
// Driver program to test above functions
int main()
{
    for (int i = 1; i < 16; i++)
        find(i);
    return 0;
}

Java

// Java program to find n'th number in a number system with
// only 3 and 4
import java.io.*;
 
class GFG {
    // Function to find n'th number in a number system with
    // only 3 and 4
    static void find(int n)
    {
        // An array of strings to store first n numbers.
        // arr[i] stores i'th number
        String[] arr = new String[n + 1];
 
        // arr[0] stores the empty string (String with 0
        // digits)
        arr[0] = "";
 
        // size indicates number of current elements in
        // arr[], m indicates number of elements added to
        // arr[] in previous iteration
        int size = 1, m = 1;
 
        // Every iteration of following loop generates and
        // adds 2*m numbers to arr[] using the m numbers
        // generated in previous iteration
        while (size <= n) {
            // Consider all numbers added in previous
            // iteration, add a prefix "3" to them and add
            // new numbers to arr[]
            for (int i = 0; i < m && (size + i) <= n; i++)
                arr[size + i] = "3" + arr[size - m + i];
 
            // Add prefix "4" to numbers of previous
            // iteration and add new numbers to arr[]
            for (int i = 0; i < m && (size + m + i) <= n;
                 i++)
                arr[size + m + i] = "4" + arr[size - m + i];
 
            // Update no. of elements added in previous
            // iteration
            m = m << 1; // Or m = m*2;
 
            // Update size
            size = size + m;
        }
        System.out.println(arr[n]);
    }
 
    // Driver program
    public static void main(String[] args)
    {
        for (int i = 0; i < 16; i++)
            find(i);
    }
}
 
// Contributed by Pramod Kumar

Python3

# Python3 program to find n'th
# number in a number system
# with only 3 and 4
 
# Function to find n'th number in a
# number system with only 3 and 4
def find(n):
     
    # An array of strings to store
    # first n numbers. arr[i] stores
    # i'th number
    arr = [''] * (n + 1);
     
    # arr[0] = ""; # arr[0] stores
    # the empty string (String with 0 digits)
 
    # size indicates number of current
    # elements in arr[]. m indicates
    # number of elements added to arr[]
    # in previous iteration.
    size = 1;
    m = 1;
 
    # Every iteration of following
    # loop generates and adds 2*m
    # numbers to arr[] using the m
    # numbers generated in previous
    # iteration.
    while (size <= n):
         
        # Consider all numbers added
        # in previous iteration, add
        # a prefix "3" to them and
        # add new numbers to arr[]
        i = 0;
        while(i < m and (size + i) <= n):
            arr[size + i] = "3" + arr[size - m + i];
            i += 1;
 
        # Add prefix "4" to numbers of
        # previous iteration and add
        # new numbers to arr[]
        i = 0;
        while(i < m and (size + m + i) <= n):
            arr[size + m + i] = "4" + arr[size - m + i];
            i += 1;
 
        # Update no. of elements added
        # in previous iteration
        m = m << 1; # Or m = m*2;
 
        # Update size
        size = size + m;
    print(arr[n]);
 
# Driver Code
for i in range(1, 16):
    find(i);
 
# This code is contributed by mits

C#

// C# program to find n'th number in a
// number system with only 3 and 4
using System;
 
class GFG {
     
    // Function to find n'th number in a
    // number system with only 3 and 4
    static void find(int n)
    {
         
        // An array of strings to store first
        // n numbers. arr[i] stores i'th number
        String[] arr = new String[n + 1];
         
        // arr[0] stores the empty string
        // (String with 0 digits)
        arr[0] = "";
 
        // size indicates number of current
        // elements in arr[], m indicates
        // number of elements added to arr[]
        // in previous iteration
        int size = 1, m = 1;
 
        // Every iteration of following loop
        // generates and adds 2*m numbers to
        // arr[] using the m numbers generated
        // in previous iteration
        while (size <= n)
        {
            // Consider all numbers added in
            // previous iteration, add a prefix
            // "3" to them and add new numbers
            // to arr[]
            for (int i = 0; i < m &&
                             (size + i) <= n; i++)
                              
                arr[size + i] = "3" +
                               arr[size - m + i];
 
            // Add prefix "4" to numbers of
            // previous iteration and add new
            // numbers to arr[]
            for (int i = 0; i < m &&
                          (size + m + i) <= n; i++)
                           
                arr[size + m + i] = "4" +
                                  arr[size - m + i];
 
            // Update no. of elements added
            // in previous iteration
            m = m << 1; // Or m = m*2;
 
            // Update size
            size = size + m;
        }
         
        Console.WriteLine(arr[n]);
    }
     
    // Driver program
    public static void Main ()
    {
        for (int i = 0; i < 16; i++)
            find(i);
    }
}
 
// This code is contributed by Sam007.

Javascript

// javascript program to find n'th number in a number system with
// only 3 and 4
 
   // Function to find n'th number in a number system with
    // only 3 and 4
function find(n)
{
 
    // An array of strings to store first n numbers.
    // arr[i] stores i'th number
    var arr = Array.from({length: n + 1}, (_, i) => " ");
 
    // arr[0] stores the empty string (String with 0
    // digits)
    arr[0] = "";
 
    // size indicates number of current elements in
    // arr, m indicates number of elements added to
    // arr in previous iteration
    var size = 1, m = 1;
 
    // Every iteration of following loop generates and
    // adds 2*m numbers to arr using the m numbers
    // generated in previous iteration
    while (size <= n)
    {
     
        // Consider all numbers added in previous
        // iteration, add a prefix "3" to them and add
        // new numbers to arr
        for (var i = 0; i < m && (size + i) <= n; i++)
            arr[size + i] = "3" + arr[size - m + i];
 
        // Add prefix "4" to numbers of previous
        // iteration and add new numbers to arr
        for (var i = 0; i < m && (size + m + i) <= n;
             i++)
            arr[size + m + i] = "4" + arr[size - m + i];
 
        // Update no. of elements added in previous
        // iteration
        m = m << 1; // Or m = m*2;
 
        // Update size
        size = size + m;
    }
    document.write(arr[n]+"<br>");
}
 
// Driver program
for (i = 0; i < 16; i++)
    find(i);
 
// This code is contributed by Amit Katiyar

输出：

3
4
33
34
43
44
333
334
343
344
433
434
443
444
3333

更好的方法(使用位)：

这个想法由Arjun J(https://auth.geeksforgeeks.org/user/camsboyfriend/profile)提出。

这里的想法是，由于我们将只处理两个数字，即3和4，因此我们可以将它们与二进制数字进行比较。

解释 ：

1)  3   -  0     (0)
2)  4   -  1     (1)

3)  33  -  00    (0)
4)  34  -  01    (1)
5)  43  -  10    (2)  
6)  44  -  11    (3)

7)  333 -  000   (0)
8)  334 -  001   (1)
9)  343 -  010   (2) 
10) 344 -  011   (3)
11) 433 -  100   (4)
12) 434 -  101   (5)
13) 443 -  110   (6)
14) 444 -  111   (7)
15) 3333 - 1000  (8)

Here we can note that 

1. Every (n – 1)’th number gets a new digit where n is a power of 2
2. Whenever a new digit is added we start the counting binary numbers from 0.
3. 0 in binary form corresponds to 3 in our number system and similarly 1  corresponds to 4.

下面是相同的C++实现：

C++

// CPP program for the above approach
#include 
using namespace std;
 
// function to find highest power of 2
// less than or equal to n
int highestPowerof2(unsigned int n)
{
    if (n < 1)
        return 0;
 
    int res = 1;
 
    for (int i = 0; i < 8 * sizeof(unsigned int); i++) {
        int curr = 1 << i;
 
        if (curr > n)
            break;
 
        res = curr;
    }
 
    return res;
}
 
// function to convert decimal to binary form
vector decToBinary(int n, int size)
{
    vector binaryNum(size + 1);
 
    int i = 0;
    while (n > 0) {
        binaryNum[i] = n % 2;
        n = (n >> 1);
        i++;
    }
 
    return binaryNum;
}
 
// Driver Code
signed main()
{
    for (int n = 1; n < 16; n++) {
        int hp2 = highestPowerof2(n + 1);
 
        int howMany = n - hp2 + 1;
 
        vector arr
            = decToBinary(howMany, log2(hp2 - 1));
 
        for (int i = log2(hp2 - 1); i >= 0; i--) {
            if (arr[i])
                cout << 4;
            else
                cout << 3;
        }
        cout << '\n';
    }
}

输出：

3
4
33
34
43
44
333
334
343
344
433
434
443
444
3333