从0到N的数字的位差总和

给定数字N ，任务是为从0到N的每个连续数字计算二进制表示形式中相应不同位的总数。

例子：

Input: N = 5
Output: 8
Explanation:
Binary Representation of numbers are:
0 -> 000,
1 -> 001,
2 -> 010,
3 -> 011,
4 -> 100,
5 -> 101
Between 1 and 0 -> 1 bit is different
Between 2 and 1 -> 2 bits are different
Between 3 and 2 -> 1 bit is different
Between 4 and 3 -> 3 bits are different
Between 5 and 4 -> 1 bit is different
Total = 1 + 2 + 1 + 3 + 1 = 8

Input: N = 11
Output: 19

天真的方法：这个想法是从0到N进行迭代，对于每对连续的元素，使用本文讨论的方法为每对元素找到不同的位数。

时间复杂度： O(N * log N)
辅助空间： (1)

高效的方法：对于高效的方法，我们必须注意以下几点：

number:   1 2 3 4 5 6  7  8
bit_diff: 1 2 1 3 1 2  1  4
sum_diff: 1 3 4 7 8 10 11 15

通过以上计算，我们可以观察到以下内容：

1. 如果N是2的理想幂，则从0到N的相应不同比特的总和由下式给出：

   sum_diff = (2^x+1-1)
   where x = log2(N)

2. 如果N不是2的完美幂，则可以将其表示为2的完美幂：

   N = 2^a + 2^b + … + 2^x

   因此，从0到N的相应不同比特的总和可以计算为：

   sum_diff = (2^a+1-1) + (2^b+1-1) + … + (2^x+1-1).

举些例子：

If N = 8, then
The binary representation of 8 is “1000”
Hence 11 = 2³
total count = (2³⁺¹ – 1)
total count = 8 – 1 = 7.

If N = 11, then
The binary representation of 11 is “1011”
Hence 11 = 2³ + 2¹ + 2⁰
=> total count = (2³⁺¹ – 1) + (2¹⁺¹ – 1) + (2⁰⁺¹ – 1)
=> total count = 15 + 3 + 1 = 19.

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to implement fast
// exponentiation
int binpow(int a, int b)
{
    int res = 1;
  
    while (b) {
        if (b & 1)
            res = res * a;
        a = a * a;
        b /= 2;
    }
    return res;
}
  
// Function to return the value
// for powers of 2
int find(int x)
{
    if (x == 0)
        return 0;
    int p = log2(x);
    return binpow(2, p + 1) - 1;
}
  
// Function to convert N into binary
string getBinary(int n)
{
    // To store binary representation
    string ans = "";
  
    // Iterate each digit of n
    while (n) {
        int dig = n % 2;
        ans += to_string(dig);
        n /= 2;
    }
  
    // Return binary representation
    return ans;
}
  
// Function to find difference in bits
int totalCountDifference(int n)
{
    // Get binary representation
    string ans = getBinary(n);
  
    // total number of bit
    // differences from 0 to N
    int req = 0;
  
    // Iterate over each binary bit
    for (int i = 0; i < ans.size(); i++) {
  
        // If current bit is '1' then add
        // the count of current bit
        if (ans[i] == '1') {
  
            req += find(binpow(2, i));
        }
    }
    return req;
}
  
// Driver Code
int main()
{
    // Given Number
    int N = 5;
  
    // Function Call
    cout << totalCountDifference(N);
    return 0;
}

Java

// Java program for the above approach 
import java.io.*; 
import java.lang.Math; 
  
class GFG{ 
      
// Function to implement fast 
// exponentiation 
static int binpow(int a, int b) 
{ 
    int res = 1; 
  
    while (b > 0)
    { 
        if (b % 2 == 1) 
            res = res * a; 
        a = a * a; 
        b /= 2; 
    } 
    return res; 
} 
  
// Function to return the 
// value for powers of 2 
static int find(int x) 
{ 
    if (x == 0) 
        return 0; 
          
    int p = (int)(Math.log(x) / Math.log(2)); 
    return binpow(2, p + 1) - 1; 
} 
  
// Function to convert N into binary 
static String getBinary(int n) 
{ 
      
    // To store the binary representation 
    String ans = ""; 
  
    // Iterate each digit of n 
    while (n > 0)
    { 
        int dig = n % 2; 
        ans += dig; 
        n /= 2; 
    } 
  
    // Return binary representation 
    return ans; 
} 
  
// Function to find difference in bits 
static int totalCountDifference(int n) 
{ 
      
    // Get binary representation 
    String ans = getBinary(n); 
  
    // total number of bit 
    // differences from 0 to N 
    int req = 0; 
  
    // Iterate over each binary bit 
    for(int i = 0; i < ans.length(); i++) 
    { 
         
       // If current bit is '1' then add 
       // the count of current bit 
       if (ans.charAt(i) == '1')
       { 
           req += find(binpow(2, i)); 
       } 
    } 
    return req; 
} 
  
// Driver code
public static void main (String[] args) 
{ 
    // Given number 
    int n = 5; 
      
    System.out.print(totalCountDifference(n)); 
} 
} 
  
// This code is contributed by spp____

Python3

# Python3 program for the above approach
from math import log
  
# Function to implement fast
# exponentiation
def binpow(a, b):
      
    res = 1
    while (b > 0):
        if (b % 2 == 1):
            res = res * a
        a = a * a
        b //= 2
    return res
  
# Function to return the value
# for powers of 2
def find(x):
      
    if (x == 0):
        return 0
    p = log(x) / log(2)
    return binpow(2, p + 1) - 1
  
# Function to convert N into binary
def getBinary(n):
      
    # To store binary representation
    ans = ""
      
    # Iterate each digit of n
    while (n > 0):
        dig = n % 2
        ans += str(dig)
        n //= 2
      
    # Return binary representation
    return ans
  
# Function to find difference in bits
def totalCountDifference(n):
      
    # Get binary representation
    ans = getBinary(n)
      
    # total number of bit
    # differences from 0 to N
    req = 0
      
    # Iterate over each binary bit
    for i in range(len(ans)):
          
        # If current bit is '1' then add
        # the count of current bit
        if (ans[i] == '1'):
            req += find(binpow(2, i))
    return req
  
# Driver Code
  
# Given Number
N = 5
  
# Function Call
print(totalCountDifference(N))
  
# This code is contributed by shubhamsingh10

C#

// C# program for the above approach 
using System;
class GFG{ 
      
// Function to implement fast 
// exponentiation 
static int binpow(int a, int b) 
{ 
    int res = 1; 
  
    while (b > 0)
    { 
        if (b % 2 == 1) 
            res = res * a; 
        a = a * a; 
        b /= 2; 
    } 
    return res; 
} 
  
// Function to return the 
// value for powers of 2 
static int find(int x) 
{ 
    if (x == 0) 
        return 0; 
          
    int p = (int)(Math.Log(x) / Math.Log(2)); 
    return binpow(2, p + 1) - 1; 
} 
  
// Function to convert N into binary 
static String getBinary(int n) 
{ 
      
    // To store the binary representation 
    String ans = ""; 
  
    // Iterate each digit of n 
    while (n > 0)
    { 
        int dig = n % 2; 
        ans += dig; 
        n /= 2; 
    } 
  
    // Return binary representation 
    return ans; 
} 
  
// Function to find difference in bits 
static int totalCountDifference(int n) 
{ 
      
    // Get binary representation 
    string ans = getBinary(n); 
  
    // total number of bit 
    // differences from 0 to N 
    int req = 0; 
  
    // Iterate over each binary bit 
    for(int i = 0; i < ans.Length; i++) 
    { 
         
       // If current bit is '1' then add 
       // the count of current bit 
       if (ans[i] == '1')
       { 
           req += find(binpow(2, i)); 
       } 
    } 
    return req; 
} 
  
// Driver code
public static void Main() 
{ 
    // Given number 
    int n = 5; 
      
    Console.Write(totalCountDifference(n)); 
} 
} 
  
// This code is contributed by Nidhi_biet

输出：

8

时间复杂度： O((log N) ² )
辅助空间： (1)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。