给定一个数组nrr [] ，该数组arr []由N个正数和形式为[L，R]的Q个查询组成，任务是在每个查询的子数组[L，R]中查找为2的幂的元素数。
例子：
Input: arr[] = { 3, 8, 5, 2, 5, 10 }, Q = {{0, 4}, {3, 5}}
Output:
2
1
Explanation:
For Query 1, the subarray [3, 8, 5, 2, 5] has 2 elements which are a power of two, 8 and 2.
For Query 2, the subarray {2, 5, 10} has 1 element which are a power of two, 2.
Input: arr[] = { 1, 2, 3, 4, 5, 6 }, Q = {{0, 4}, {1, 5}}
Output:
3
2
朴素的方法：为解决上述问题，朴素的方法是对于所有Q查询，我们可以遍历数组中的每个L和R，并找到子数组中为2的幂的元素数[L，R ]。
时间复杂度： O(N * Q)
高效方法：
为了优化上述方法，这里的想法是使用前缀和数组。
- 最初，前缀总和数组的所有索引都包含0。
- 如果当前数组元素是2的幂，则遍历给定数组并将此索引的前缀数组设置为1，否则将其保留为0。
- 现在，通过添加前一个索引前缀数组值来计算当前索引的前缀总和，从而获得前缀总和。 prefix [i]将存储从1到i为2的幂的元素的数量。
- 一旦有了前缀数组，我们只需要为每个查询返回prefix [r] – prefix [l-1] 。
下面是上述方法的实现，
C++
// C++ implementation to find
// elements that are a power of two
#include
using namespace std;
const int MAX = 10000;
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
int prefix[MAX + 1];
bool isPowerOfTwo(int x)
{
if (x && (!(x & (x - 1))))
return true;
return false;
}
// Function to find the maximum range
// whose sum is divisible by M.
void computePrefix(int n, int a[])
{
// Calculate the prefix sum
if (isPowerOfTwo(a[0]))
prefix[0] = 1;
for (int i = 1; i < n; i++) {
prefix[i] = prefix[i - 1];
if (isPowerOfTwo(a[i]))
prefix[i]++;
}
}
// Function to return the number of elements
// which are a power of two in a subarray
int query(int L, int R)
{
return prefix[R] - prefix[L - 1];
}
// Driver code
int main()
{
int A[] = { 3, 8, 5, 2, 5, 10 };
int N = sizeof(A) / sizeof(A[0]);
int Q = 2;
computePrefix(N, A);
cout << query(0, 4) << "\n";
cout << query(3, 5) << "\n";
return 0;
} Java
// Java implementation to find
// elements that are a power of two
import java.util.*;
class GFG{
static final int MAX = 10000;
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
static int[] prefix = new int[MAX + 1];
static boolean isPowerOfTwo(int x)
{
if (x != 0 && ((x & (x - 1)) == 0))
return true;
return false;
}
// Function to find the maximum range
// whose sum is divisible by M.
static void computePrefix(int n, int a[])
{
// Calculate the prefix sum
if (isPowerOfTwo(a[0]))
prefix[0] = 1;
for (int i = 1; i < n; i++)
{
prefix[i] = prefix[i - 1];
if (isPowerOfTwo(a[i]))
prefix[i]++;
}
}
// Function to return the number of elements
// which are a power of two in a subarray
static int query(int L, int R)
{
if (L == 0)
return prefix[R];
return prefix[R] - prefix[L - 1];
}
// Driver code
public static void main(String[] args)
{
int A[] = { 3, 8, 5, 2, 5, 10 };
int N = A.length;
int Q = 2;
computePrefix(N, A);
System.out.println(query(0, 4));
System.out.println(query(3, 5));
}
}
// This code is contributed by offbeatPython3
# Python3 implementation to find
# elements that are a power of two
MAX = 10000
# prefix[i] is going to store the
# number of elements which are a
# power of two till i (including i).
prefix = [0] * (MAX + 1)
def isPowerOfTwo(x):
if (x and (not (x & (x - 1)))):
return True
return False
# Function to find the maximum range
# whose sum is divisible by M.
def computePrefix(n, a):
# Calculate the prefix sum
if (isPowerOfTwo(a[0])):
prefix[0] = 1
for i in range(1, n):
prefix[i] = prefix[i - 1]
if (isPowerOfTwo(a[i])):
prefix[i] += 1
# Function to return the number of elements
# which are a power of two in a subarray
def query(L, R):
return prefix[R] - prefix[L - 1]
# Driver code
if __name__ == "__main__":
A = [ 3, 8, 5, 2, 5, 10 ]
N = len(A)
Q = 2
computePrefix(N, A)
print(query(0, 4))
print(query(3, 5))
# This code is contributed by chitranayalC#
// C# implementation to find
// elements that are a power of two
using System;
class GFG{
static int MAX = 10000;
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
static int[] prefix = new int[MAX + 1];
static bool isPowerOfTwo(int x)
{
if (x != 0 && ((x & (x - 1)) == 0))
return true;
return false;
}
// Function to find the maximum range
// whose sum is divisible by M.
static void computePrefix(int n, int []a)
{
// Calculate the prefix sum
if (isPowerOfTwo(a[0]))
prefix[0] = 1;
for (int i = 1; i < n; i++)
{
prefix[i] = prefix[i - 1];
if (isPowerOfTwo(a[i]))
prefix[i]++;
}
}
// Function to return the number of elements
// which are a power of two in a subarray
static int query(int L, int R)
{
if (L == 0)
return prefix[R];
return prefix[R] - prefix[L - 1];
}
// Driver code
public static void Main()
{
int []A = { 3, 8, 5, 2, 5, 10 };
int N = A.Length;
computePrefix(N, A);
Console.WriteLine(query(0, 4));
Console.WriteLine(query(3, 5));
}
}
// This code is contributed by Code_Mech输出：
2
1
时间复杂度： O(max(Q，N))