查找 L 到 R 中的数字，这与提高到 setbit 计数的数字总和相同

给定一个数字范围[L, R] ，任务是找到给定范围内的所有数字X使得X = 数字之和提升到 X 的 setbit 计数，即，如果在 X 和 X 的二进制表示中有N个 setbits = x ₁ x ₂ x ₃ … 然后 X = (x ₁ ) ^N + (x ₂ ) ^N + (x ₃ ) ^N + 。 . .

例子：

Input: L = 0, R = 10000
Output: 1, 2, 4, 8, 4150, 9474
Explanation: For 2 (binary = 10): setbit count = 1. and 2 = 2^1.
So this is a required number. Same for the other numbers also.

Input: L = 10000, R = 1000000
Output: -1
Explanation: There are no such numbers in the given range.

方法：给定的问题可以通过检查 [L, R] 范围内的所有数字以及它们是否满足条件来解决。这可以在Brian Kernighan 的算法的帮助下完成。

请按照以下步骤解决问题。

• 运行从L到R的循环，并在每次迭代中检查数字是否为索引号。
• 首先从其二进制表示中计算十进制数中的设置位数。
• 然后， Initialize Original = N , Res = 0 , Index = Count of Set bits 。
• 在 N > 0 时运行一个循环
  • 从数字中找到最后一位数字说（ L ），
  • 将L^索引添加到Res。
  • 从号码中删除最后一位数字。
• 如果Original = Res ，这将是所需数字之一。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
#define ll long long
 
// Function to return Number of
// set bits in any decimal number
int countSetBits(ll N)
{
    int Count = 0;
    while (N) {
        N = N & (N - 1);
        Count++;
    }
    return Count;
}
 
// Function to check whether the
// number is index number or not
bool check(int Index, ll N)
{
    ll Original = N, Res = 0;
   
    if(N == 0)
      return false;
   
    while (N != 0) {
        int L = N % 10;
        Res += pow(L, Index);
        N = N / 10;
    }
    return Original == Res;
}
 
// Function to find the numbers
vector findNum(int l, int r)
{
    // Vector to store the numbers
    vector ans;
 
    for (ll i = l; i <= r; i++) {
        int BitCount = countSetBits(i);
        if (check(BitCount, i))
            ans.push_back(i);
    }
    return ans;
}
 
// Driver Code
int main()
{
    int L = 0, R = 10000;
 
    // Function call
    vector res = findNum(L, R);
           
    if(res.size()==0)
        cout << -1 << endl;
   
    for (int x : res)
        cout << x << " ";
   
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
  // Function to return Number of
  // set bits in any decimal number
  static int countSetBits(long N)
  {
    int Count = 0;
    while (N != 0) {
      N = N & (N - 1);
      Count++;
    }
    return Count;
  }
 
  // Function to check whether the
  // number is index number or not
  static boolean check(int Index, long N)
  {
    long Original = N, Res = 0;
 
    if(N == 0)
      return false;
 
    while (N != 0) {
      long L = N % 10;
      Res += Math.pow(L, Index);
      N = N / 10;
    }
    return Original == Res;
  }
 
  // Function to find the numbers
  static Vector findNum(int l, int r)
  {
 
    // Vector to store the numbers
    Vector ans = new Vector();
 
    for (int i = l; i <= r; i++) {
      int BitCount = countSetBits(i);
      if (check(BitCount, i))
        ans.add(i);
    }
    return ans;
  }
 
  // Driver Code
  public static void main (String[] args) {   
    int L = 0, R = 10000;
 
    // Function call
    Vector res = findNum(L, R);
 
    if(res.size()==0)
      System.out.println(-1);
 
    res.forEach((x) -> System.out.print(x + " "));
  }
}
 
// This code is contributed by hrithikgarg03188.

Javascript

    // JavaScript program for the above approach
 
 
    // Function to return Number of
    // set bits in any decimal number
    const countSetBits = (N) => {
        let Count = 0;
        while (N) {
            N = N & (N - 1);
            Count++;
        }
        return Count;
    }
 
    // Function to check whether the
    // number is index number or not
    const check = (Index, N) => {
        let Original = N, Res = 0;
        while (N != 0) {
            let L = N % 10;
            Res += Math.pow(L, Index);
            N = parseInt(N / 10);
        }
        return Original == Res;
    }
 
    // Function to find the numbers
    const findNum = (l, r) => {
        // Vector to store the numbers
        let ans = [];
 
        for (let i = l; i <= r; i++) {
            let BitCount = countSetBits(i);
            if (check(BitCount, i))
                ans.push(i);
        }
        return ans;
    }
 
    // Driver Code
 
    let L = 0, R = 10000;
 
    // Function call
    let res = findNum(L, R);
    for (let x in res)
        document.write(`${res[x]} `);
 
// This code is contributed by rakeshsahni
 

输出

1 2 4 8 4150 9474 

时间复杂度： O(R * d) 其中d是数字中的最大位数
辅助空间： O(1)