给定正整数数组和包含两个整数的q查询，L＆R。任务是查找给定范围内的置位位数。
先决条件：按位黑客
例子 ：
Input : Arr[] = { 1, 5, 6, 10, 9, 4 }
Query : 2
L & R
1 5
2 4
Output : 9
6
Input : Arr[] = { 1, 10, 5, 2, 8, 11, 15 }
Query : 2
L & R
2 4
1 5
Output : 4
9解决此问题的简单方法是运行一个从L到R的循环，并对范围中的设置位数进行计数。此解决方案为每个查询取O(nlog(s))(其中s是位大小)。
高效的解决方案基于这样一个事实：如果我们将所有数字的所有设置位的计数存储在“ BitCounts”数组中，那么我们将在O(1)时间内回答每个查询。因此，开始遍历数组的元素并计算每个元素的设置位并存储在数组中。现在，找到该数组的累加和。该数组将有助于回答查询。
BitCount[] that will store the count of set bits
in a number.
Run a Loop from 0 to 31 "for 32 bits size integer "
-> mark elements with i'th bit set
Run an inner Loop from 0 to size of Array "Arr"
-> Check whether the current bit is set or not
-> if it's set then mark it.
long temp = arr[j] >> i;
if (temp %2 != 0)
BitCount[j] += 1以下是上述想法的实现。
C++
// C++ program to Range query for
// Count number of set bits
#include
using namespace std;
// 2-D array that will stored the count
// of bits set in element of array
int BitCount[10000] = { 0 };
// Function store the set bit
// count in BitCount Array
void fillSetBitsMatrix(int arr[], int n)
{
// traverse over all bits
for (int i = 0; i < 32; i++) {
// mark elements with i'th bit set
for (int j = 0; j < n; j++) {
// Check whether the current bit is
// set or not if it's set then mark it.
long temp = arr[j] >> i;
if (temp % 2 != 0)
BitCount[j] += 1;
}
}
// store cumulative sum of bits
for (int i = 1; i < n; i++)
BitCount[i] += BitCount[i - 1];
}
// Function to process queries
void Query(int Q[][2], int q)
{
for (int i = 0; i < q; i++)
cout << (BitCount[Q[i][1]] -
BitCount[Q[i][0] - 1]) << endl;
}
// Driver Code
int main()
{
int Arr[] = { 1, 5, 6, 10, 9, 4, 67 };
int n = sizeof(Arr) / sizeof(Arr[0]);
fillSetBitsMatrix(Arr, n);
int q = 2;
int Q[2][2] = { { 1, 5 }, { 2, 6 } };
Query(Q, q);
return 0;
} Java
// Java program to Range query for
// Count number of set bits
import java.io.*;
class GFG {
// 2-D array that will stored the count
// of bits set in element of array
static int BitCount[] = new int[10000];
// Function store the set bit
// count in BitCount Array
static void fillSetBitsMatrix(int arr[], int n)
{
// traverse over all bits
for (int i = 0; i < 32; i++) {
// mark elements with i'th bit set
for (int j = 0; j < n; j++) {
// Check whether the current
// bit is set or not if it's
// set then mark it.
long temp = arr[j] >> i;
if (temp % 2 != 0)
BitCount[j] += 1;
}
}
// store cumulative sum of bits
for (int i = 1; i < n; i++)
BitCount[i] += BitCount[i - 1];
}
// Function to process queries
static void Query(int Q[][], int q)
{
for (int i = 0; i < q; i++)
System.out.println( (BitCount[Q[i][1]]
- BitCount[Q[i][0] - 1]));
}
// Driver Code
public static void main (String[] args)
{
int Arr[] = { 1, 5, 6, 10, 9, 4, 67 };
int n = Arr.length;
fillSetBitsMatrix(Arr, n);
int q = 2;
int Q[][] = { { 1, 5 }, { 2, 6 } };
Query(Q, q);
}
}
// This code is contributed by anuj_67.Python3
# Python3 program to Range query for
# Count number of set bits
# 2-D array that will stored the count
# of bits set in element of array
BitCount = [0] * 10000
# Function store the set bit
# count in BitCount Array
def fillSetBitsmatrix(arr: list, n: int):
global BitCount
# traverse over all bits
for i in range(32):
# mark elements with i'th bit set
for j in range(n):
# Check whether the current bit is
# set or not if it's set then mark it.
temp = arr[j] >> i
if temp % 2 != 0:
BitCount[j] += 1
# store cumulative sum of bits
for i in range(1, n):
BitCount[i] += BitCount[i - 1]
# Function to process queries
def Query(Q: list, q: int):
for i in range(q):
print(BitCount[Q[i][1]] - BitCount[Q[i][0] - 1])
# Driver Code
if __name__ == "__main__":
Arr = [1, 5, 6, 10, 9, 4, 67]
n = len(Arr)
fillSetBitsmatrix(Arr, n)
q = 2
Q = [(1, 5), (2, 6)]
Query(Q, q)
# This code is contributed by
# sanjeev2552C#
// C# program to Range query for
// Count number of set bits
using System;
class GFG {
// 2-D array that will stored the count
// of bits set in element of array
static int []BitCount = new int[10000];
// Function store the set bit
// count in BitCount Array
static void fillSetBitsMatrix(int []arr, int n)
{
// traverse over all bits
for (int i = 0; i < 32; i++) {
// mark elements with i'th bit set
for (int j = 0; j < n; j++) {
// Check whether the current
// bit is set or not if it's
// set then mark it.
long temp = arr[j] >> i;
if (temp % 2 != 0)
BitCount[j] += 1;
}
}
// store cumulative sum of bits
for (int i = 1; i < n; i++)
BitCount[i] += BitCount[i - 1];
}
// Function to process queries
static void Query(int [,]Q, int q)
{
for (int i = 0; i < q; i++)
Console.WriteLine( (BitCount[Q[i,1]]
- BitCount[Q[i,0] - 1]));
}
// Driver Code
public static void Main ()
{
int []Arr = { 1, 5, 6, 10, 9, 4, 67 };
int n = Arr.Length;
fillSetBitsMatrix(Arr, n);
int q = 2;
int [,]Q = { { 1, 5 }, { 2, 6 } };
Query(Q, q);
}
}
// This code is contributed by anuj_67.Javascript
输出：
9
10时间复杂度：每个查询为O(1)。