在给定数组中为 Q 查询计算索引范围 [L, R] 中的设置位

给定一个包含N个整数的数组arr[]和一个包含{L, R}形式的Q个查询的数组queries[] ，任务是计算每个查询在数组arr中从L到R的设置位总数。

例子：

Input: arr[]={1, 2, 3, 4, 5, 6}, queries[]={{0, 2}, {1, 1}, {3, 5}}
Output:
4
2
5
Explanation:
Query 1 (0, 2): Elements {1, 2, 3} have set bits {1, 1, 2} respectively. So sum=1+1+2=4
Query 1 (1, 1): Element {2} have set bit {1}. So sum=1
Query 1 (3, 5): Elements {4, 5, 6} have set bits {1, 2, 2} respectively. So sum=1+2+2=5

Input: arr[]={2, 4, 3, 5, 6}, queries[]={{0, 1}, {2, 4}}
Output:
2
6

方法：要解决这个问题，请按照以下步骤操作：

1. 创建一个向量，比如onBits ，其中每个ⁱ元素将表示从arr[0]到arr[i]的设置位数。
2. 从i=0 到 i运行一个循环，并在每次迭代中，找到arr[i]中设置的位数，比如 bits 并制作onBits[i]=onBits[i-1] + bits 。
3. 现在遍历数组查询，对于每个查询{L, R} ，从L到R的设置位数为onBits[R]-onBits[L-1] 。
4. 根据以上观察打印答案。

下面是上述方法的实现：

C++

// C++ code for the above approach
 
#include 
using namespace std;
 
// Function to return the number
// of set bits from L to R
int setBitsinLtoR(int L, int R,
                  vector& onBits)
{
    if (L == 0) {
        return onBits[R];
    }
 
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
void totalSetBits(vector& arr,
                  vector >& queries)
{
    int N = arr.size();
    vector onBits(N, 0);
    onBits[0] = __builtin_popcount(arr[0]);
 
    for (int i = 1; i < N; ++i) {
        onBits[i]
            = onBits[i - 1]
              + __builtin_popcount(arr[i]);
    }
 
    // Finding for Q queries
    for (auto x : queries) {
        int L = x.first;
        int R = x.second;
 
        cout << setBitsinLtoR(L, R, onBits)
             << endl;
    }
}
 
// Driver Code
int main()
{
    vector arr = { 1, 2, 3, 4, 5, 6 };
    vector > queries
        = { { 0, 2 }, { 1, 1 }, { 3, 5 } };
 
    totalSetBits(arr, queries);
}

---

Java

// Java code for the above approach
import java.util.*;
class GFG{
 
static class pair
{
    int first, second;
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
   
// Function to return the number
// of set bits from L to R
static int setBitsinLtoR(int L, int R,int []onBits)
{
    if (L == 0) {
        return onBits[R];
    }
 
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
static void totalSetBits(int[] arr,
                  pair[] queries)
{
    int N = arr.length;
     
    int []onBits = new int[N];
    onBits[0] = Integer.bitCount(arr[0]);
 
    for (int i = 1; i < N; ++i) {
        onBits[i]
            = onBits[i - 1]
              + Integer.bitCount(arr[i]);
    }
 
    // Finding for Q queries
    for (pair x : queries) {
        int L = x.first;
        int R = x.second;
 
        System.out.print(setBitsinLtoR(L, R, onBits)
             +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    pair []queries
        = { new pair( 0, 2 ), new pair( 1, 1 ), new pair( 3, 5 ) };
 
    totalSetBits(arr, queries);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
def  __builtin_popcount(n):
    count = 0
    while (n):
        count += n & 1
        n >>= 1
    return count
 
# Function to return the number
# of set bits from L to R
def setBitsinLtoR(L, R, onBits):
    if (L == 0):
        return onBits[R];
    return onBits[R] - onBits[L - 1];
 
# Function to find the number
# of set bits for Q queries
def totalSetBits(arr, queries):
    N = len(arr);
    onBits = [0] * N
    onBits[0] = __builtin_popcount(arr[0]);
 
    for i in range(1, N):
        onBits[i] = onBits[i - 1] +  __builtin_popcount(arr[i]);
 
    # Finding for Q queries
    for x in queries:
        L = x[0];
        R = x[1];
 
        print(setBitsinLtoR(L, R, onBits))
 
# Driver Code
arr = [ 1, 2, 3, 4, 5, 6 ];
queries = [ [ 0, 2 ], [ 1, 1 ], [ 3, 5 ] ];
 
totalSetBits(arr, queries);
 
# This code is contributed by gfgking

C#

// C# code for the above approach
using System;
 
class GFG{
 
class pair
{
    public int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
   
// Function to return the number
// of set bits from L to R
static int setBitsinLtoR(int L, int R, int []onBits)
{
    if (L == 0)
    {
        return onBits[R];
    }
     
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
static void totalSetBits(int[] arr,
                         pair[] queries)
{
    int N = arr.Length;
    int []onBits = new int[N];
    onBits[0] = bitCount(arr[0]);
 
    for(int i = 1; i < N; ++i)
    {
        onBits[i] = onBits[i - 1] +
                    bitCount(arr[i]);
    }
 
    // Finding for Q queries
    foreach (pair x in queries)
    {
        int L = x.first;
        int R = x.second;
 
        Console.Write(setBitsinLtoR(L, R, onBits) +
                      "\n");
    }
}
 
static int bitCount(int x)
{
    int setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    pair []queries = { new pair(0, 2),
                       new pair(1, 1),
                       new pair(3, 5) };
 
    totalSetBits(arr, queries);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// JavaScript code for the above approach
function __builtin_popcount(x)
{
    return((x).toString(2).split('').
           filter(x => x == '1').length);
}
 
// Function to return the number
// of set bits from L to R
function setBitsinLtoR(L, R, onBits)
{
    if (L == 0)
    {
        return onBits[R];
    }
    return onBits[R] - onBits[L - 1];
}
 
// Function to find the number
// of set bits for Q queries
function totalSetBits(arr, queries)
{
    let N = arr.length;
    let onBits = new Array(N).fill(0);
    onBits[0] = __builtin_popcount(arr[0]);
 
    for(let i = 1; i < N; ++i)
    {
        onBits[i] = onBits[i - 1] +
                    __builtin_popcount(arr[i]);
    }
 
    // Finding for Q queries
    for(let x of queries)
    {
        let L = x[0];
        let R = x[1];
 
        document.write(
            setBitsinLtoR(L, R, onBits) + '<br>')
    }
}
 
// Driver Code
let arr = [ 1, 2, 3, 4, 5, 6 ];
let queries = [ [ 0, 2 ], [ 1, 1 ], [ 3, 5 ] ];
 
totalSetBits(arr, queries);
 
// This code is contributed by Potta Lokesh
 

输出

4
1
5

时间复杂度： O(Q*N*logK)，其中 N 是数组 arr 的大小，Q 是查询次数，K 是位数
辅助空间： O(N)