给定一个由N个正整数组成的数组arr [] ，任务是回答Q个查询，其中每个查询由一个范围[L，R]组成，您必须检查给定索引范围中元素的按位与是偶数还是偶数?奇怪的。
例子：
Input: arr[] = {1, 1, 2, 3}, Q[][] = {{1, 2}, {0, 1}}
Output:
Even
Odd
(1 & 2) = 0 which is even.
(1 & 1) = 1 which is odd.
Input: arr[] = {2, 1, 5, 7, 6, 8, 9}, Q[][] = {{0, 2}, {1, 2}, {2, 3}, {3, 6}}
Output:
Even
Odd
Odd
Even
方法：
- 如果索引范围[L，R]中的所有元素均是奇数，则按位与将为奇数，否则为偶数。
- 因此，请检查索引范围[L，R]中的所有元素是否为奇数。
- 为了最佳地回答每个查询，请创建一个数组并标记值为奇数的位置，然后使用此数组的前缀和。
- 现在，可以将索引范围[L，R]中的奇数计数计算为pre [R] – pre [L – 1] 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int MAXN = 1000005;
int even[MAXN], odd[MAXN];
// Function to precompute the count
// of even and odd numbers
void precompute(int arr[], int n)
{
for (int i = 0; i < n; i++) {
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (int i = 1; i < n; i++) {
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
bool isOdd(int L, int R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
int cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
void performQueries(int a[], int n, int q[][2], int m)
{
precompute(a, n);
// Perform queries
for (int i = 0; i < m; i++) {
int L = q[i][0], R = q[i][1];
if (isOdd(L, R))
cout << "Odd\n";
else
cout << "Even\n";
}
}
// Driver code
int main()
{
int a[] = { 2, 1, 5, 7, 6, 8, 9 };
int n = sizeof(a) / sizeof(a[0]);
// Queries
int q[][2] = { { 0, 2 }, { 1, 2 }, { 2, 3 }, { 3, 6 } };
int m = sizeof(q) / sizeof(q[0]);
performQueries(a, n, q, m);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static final int MAXN = 1000005;
static int even[] = new int[MAXN];
static int odd[] = new int[MAXN];
// Function to precompute the count
// of even and odd numbers
static void precompute(int arr[], int n)
{
for (int i = 0; i < n; i++)
{
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (int i = 1; i < n; i++)
{
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
static boolean isOdd(int L, int R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
int cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
static void performQueries(int a[], int n,
int q[][], int m)
{
precompute(a, n);
// Perform queries
for (int i = 0; i < m; i++)
{
int L = q[i][0], R = q[i][1];
if (isOdd(L, R))
System.out.println("Odd");
else
System.out.println("Even");
}
}
// Driver code
public static void main(String args[])
{
int []a = { 2, 1, 5, 7, 6, 8, 9 };
int n = a.length;
// Queries
int q[][] = { { 0, 2 }, { 1, 2 }, { 2, 3 }, { 3, 6 } };
int m = q.length;
performQueries(a, n, q, m);
}
}
// This code is contributed by AnkitRai01Python3
# Python implementation of the approach
MAXN = 1000005;
even = [0] * MAXN;
odd = [0] * MAXN;
# Function to precompute the count
# of even and odd numbers
def precompute(arr, n):
for i in range(n):
# If the current element is odd
# then put 1 at odd[i]
if (arr[i] % 2 == 1):
odd[i] = 1;
# If the current element is even
# then put 1 at even[i]
if (arr[i] % 2 == 0):
even[i] = 1;
# Taking the prefix sums of these two arrays
# so we can get the count of even and odd
# numbers in a range [L, R] in O(1)
for i in range(1, n):
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
# Function that returns True if the bitwise
# AND of the subarray a[L...R] is odd
def isOdd(L, R):
# cnt will store the count of odd
# numbers in the range [L, R]
cnt = odd[R];
if (L > 0):
cnt -= odd[L - 1];
# Check if all the numbers in
# the range are odd or not
if (cnt == R - L + 1):
return True;
return False;
# Function to perform the queries
def performQueries(a, n, q, m):
precompute(a, n);
# Perform queries
for i in range(m):
L = q[i][0];
R = q[i][1];
if (isOdd(L, R)):
print("Odd");
else:
print("Even");
# Driver code
if __name__ == '__main__':
a = [ 2, 1, 5, 7, 6, 8, 9 ];
n = len(a);
# Queries
q = [[ 0, 2 ],[ 1, 2 ],[ 2, 3 ],[ 3, 6 ]];
m = len(q);
performQueries(a, n, q, m);
# This code is contributed by PrinciRaj1992C#
// C# implementation of the approach
using System;
class GFG
{
static readonly int MAXN = 1000005;
static int []even = new int[MAXN];
static int []odd = new int[MAXN];
// Function to precompute the count
// of even and odd numbers
static void precompute(int []arr, int n)
{
for (int i = 0; i < n; i++)
{
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (int i = 1; i < n; i++)
{
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
static bool isOdd(int L, int R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
int cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
static void performQueries(int []a, int n,
int [,]q, int m)
{
precompute(a, n);
// Perform queries
for (int i = 0; i < m; i++)
{
int L = q[i, 0], R = q[i, 1];
if (isOdd(L, R))
Console.WriteLine("Odd");
else
Console.WriteLine("Even");
}
}
// Driver code
public static void Main(String []args)
{
int []a = { 2, 1, 5, 7, 6, 8, 9 };
int n = a.Length;
// Queries
int [,]q = { { 0, 2 }, { 1, 2 },
{ 2, 3 }, { 3, 6 } };
int m = q.GetLength(0);
performQueries(a, n, q, m);
}
}
// This code is contributed by 29AjayKumar输出：
Even
Odd
Odd
Even
时间复杂度： O(Q)，其中Q是查询数。