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📜  查询二进制数组的子数组的十进制值

📅  最后修改于: 2022-05-13 01:57:50.166000             🧑  作者: Mango

查询二进制数组的子数组的十进制值

给定一个二进制数组 arr[]，我们要找到由子数组 a[l..r] 表示的数字。有多个这样的查询。
例子： 

Input :  arr[] = {1, 0, 1, 0, 1, 1};
         l = 2, r = 4
         l = 4, r = 5
Output : 5
         3 
Subarray 2 to 4 is 101 which is 5 in decimal.
Subarray 4 to 5 is 11 which is 3 in decimal.

Input : arr[] = {1, 1, 1}
        l = 0, r = 2
        l = 1, r = 2
Output : 7
         3

一个简单的解决方案是使用简单的二进制到十进制转换来计算每个给定范围的十进制值。这里每个查询都需要 O(len) 时间，其中 len 是范围的长度。
一个有效的解决方案是进行每次计算，以便可以在 O(1) 时间内回答查询。
子数组 arr[l..r] 表示的数字是 arr[l]* 2^{r-l} + arr[l+1]* 2^{r - l - 1} ….. + arr[r]* 2^{r-r}

  1. 制作一个与给定数组大小相同的数组 pre[]，其中 pre[i] 存储 arr[j]* 的总和2^{n - 1 - j} 其中 j 包括从 i 到 n-1 的每个值。
  2. 子数组 arr[l..r] 表示的数字将等于 (pre[l] – pre[r+1])/ 2^{n-1-r} .pre[l] – pre[r+1] 等于 arr[l]* 2^{n - 1 - l} + arr[l+1]* 2^{n - 1 - l - 1} +……arr[r]* 2^{n - 1 - r} .所以如果我们把它除以2^{n - 1 - r} ，我们得到所需的答案

C++
// C++ implementation of finding number
// represented by binary subarray
#include 
using namespace std;
 
// Fills pre[]
void precompute(int arr[], int n, int pre[])
{
    memset(pre, 0, n * sizeof(int));
    pre[n - 1] = arr[n - 1] * pow(2, 0);
    for (int i = n - 2; i >= 0; i--)
        pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));
}
 
// returns the number represented by a binary
// subarray l to r
int decimalOfSubarr(int arr[], int l, int r,
                    int n, int pre[])
{
    // if r is equal to n-1 r+1 does not exist
    if (r != n - 1)
        return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));
 
    return pre[l] / (1 << (n - 1 - r));
}
 
// Driver Function
int main()
{
    int arr[] = { 1, 0, 1, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int pre[n];
    precompute(arr, n, pre);
    cout << decimalOfSubarr(arr, 2, 4, n, pre) << endl;
    cout << decimalOfSubarr(arr, 4, 5, n, pre) << endl;
    return 0;
}

Java
// Java implementation of finding number
// represented by binary subarray
import java.util.Arrays;
 
class GFG {
 
    // Fills pre[]
    static void precompute(int arr[], int n, int pre[])
    {
        Arrays.fill(pre, 0);
 
        pre[n - 1] = arr[n - 1] * (int)(Math.pow(2, 0));
        for (int i = n - 2; i >= 0; i--)
            pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));
    }
 
    // returns the number represented by a binary
    // subarray l to r
    static int decimalOfSubarr(int arr[], int l, int r,
                               int n, int pre[])
    {
 
        // if r is equal to n-1 r+1 does not exist
        if (r != n - 1)
            return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));
 
        return pre[l] / (1 << (n - 1 - r));
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 0, 1, 0, 1, 1 };
        int n = arr.length;
 
        int pre[] = new int[n];
        precompute(arr, n, pre);
 
        System.out.println(decimalOfSubarr(arr,
                                           2, 4, n, pre));
 
        System.out.println(decimalOfSubarr(arr,
                                           4, 5, n, pre));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3
# implementation of finding number
# represented by binary subarray
from math import pow
 
# Fills pre[]
def precompute(arr, n, pre):
     
    pre[n - 1] = arr[n - 1] * pow(2, 0)
    i = n - 2
    while(i >= 0):
        pre[i] = (pre[i + 1] + arr[i] *
                 (1 << (n - 1 - i)))
        i -= 1
 
# returns the number represented by
# a binary subarray l to r
def decimalOfSubarr(arr, l, r, n, pre):
     
    # if r is equal to n-1 r+1 does not exist
    if (r != n - 1):
        return ((pre[l] - pre[r + 1]) /
                (1 << (n - 1 - r)))
 
    return pre[l] / (1 << (n - 1 - r))
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 0, 1, 0, 1, 1]
    n = len(arr)
 
    pre = [0 for i in range(n)]
    precompute(arr, n, pre)
    print(int(decimalOfSubarr(arr, 2, 4, n, pre)))
    print(int(decimalOfSubarr(arr, 4, 5, n, pre)))
 
# This code is contributed by
# Surendra_Gangwar

C#
// C# implementation of finding number
// represented by binary subarray
using System;
 
class GFG {
 
    // Fills pre[]
    static void precompute(int[] arr, int n, int[] pre)
    {
        for (int i = 0; i < n; i++)
            pre[i] = 0;
 
        pre[n - 1] = arr[n - 1] * (int)(Math.Pow(2, 0));
        for (int i = n - 2; i >= 0; i--)
            pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));
    }
 
    // returns the number represented by
    // a binary subarray l to r
    static int decimalOfSubarr(int[] arr, int l, int r,
                                      int n, int[] pre)
    {
        // if r is equal to n-1 r+1 does not exist
        if (r != n - 1)
            return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));
 
        return pre[l] / (1 << (n - 1 - r));
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 0, 1, 0, 1, 1 };
        int n = arr.Length;
 
        int[] pre = new int[n];
        precompute(arr, n, pre);
 
        Console.WriteLine(decimalOfSubarr(arr,
                                        2, 4, n, pre));
 
        Console.WriteLine(decimalOfSubarr(arr,
                                        4, 5, n, pre));
    }
}
 
// This code is contributed by vt_m.

PHP
= 0; $i--)
        $pre[$i] = $pre[$i + 1] + $arr[$i] *
                           (1 << ($n - 1 - $i));
}
 
// returns the number represented by
// a binary subarray l to r
function decimalOfSubarr(&$arr, $l, $r, $n, &$pre)
{
    // if r is equal to n-1 r+1 does not exist
    if ($r != $n - 1)
        return ($pre[$l] - $pre[$r + 1]) /
                    (1 << ($n - 1 - $r));
 
    return $pre[$l] / (1 << ($n - 1 - $r));
}
 
// Driver Code
$arr = array(1, 0, 1, 0, 1, 1 );
$n = sizeof($arr);
 
$pre = array_fill(0, $n, NULL);
precompute($arr, $n, $pre);
echo decimalOfSubarr($arr, 2, 4, $n, $pre) . "\n";
echo decimalOfSubarr($arr, 4, 5, $n, $pre) . "\n";
 
// This code is contributed by ita_c
?>

Javascript

输出： 

5
3