给定N个整数的数组arr [] ，任务是执行以下两个查询：
- query(L，R) ：从L到R打印子数组中的偶数奇偶校验数字。
- update(i，x) ：将索引i的数组元素引用更新为x。
例子：
Input: arr[] = {18, 15, 8, 9, 14, 5}
Query 1: query(L = 0, R = 4)
Query 2: update(i = 3, x = 11)
Query 3: query(L = 0, R = 4)
Output:
3
2
Explanation:
Query 1: Subarray is {18, 15, 8, 9, 14}
Binary Representation of these elements –
18 => 10010, Parity = 2
15 => 1111, Parity = 4
8 => 1000, Parity = 1
9 => 1001, Parity = 2
14 => 1110, Parity = 3
Subarray[0-4] have 3 elements with even parity.
Query 2: Update arr[3] = 11
Updated array, {18, 15, 8, 11, 14, 5}
Query 3: Subarray is {18, 15, 8, 11, 14}
Binary Representation of these elements –
18 => 10010, Parity = 2
15 => 1111, Parity = 4
8 => 1000, Parity = 1
11 => 1011, Parity = 3
14 => 1110, Parity = 3
Subarray[0-4] have 2 elements with even parity.
方法：想法是使用段树查询数组范围内偶数奇偶校验元素的计数并同时更新。
我们可以通过迭代数字的二进制表示的每个位并计算设置位的数量来找到当前值的奇偶校验。然后检查奇偶校验是否为偶数。如果它具有偶校验，则将其设置为1，否则设置为0。
构建段树：
- 段树的叶节点表示为0(如果它是奇数奇偶校验数)或1(如果它是偶数奇偶校验数)。
- 段树的内部节点等于其子节点的总和，因此，一个节点表示从L到R的总偶数奇偶校验数，范围[L，R]落在该节点下，而其下的子树。
处理查询：
- Query(L，R)：每当我们从头到尾收到查询时，我们都可以在段树中查询从头到尾范围内的节点总数，这又表示范围开始范围内的偶数奇偶校验数结束。
- Update(i，x)：要执行更新查询以将索引i的值更新为x，我们检查以下情况：
- 情况1：如果先前值和新值都为偶校验数字
子数组中偶数奇偶校验数的计数不变，因此我们只更新数组而不修改段树 - 情况2：如果以前的值和新值都不都是偶数
子数组中偶数奇偶校验数的计数不变，因此我们只更新数组而不修改段树 - 情况3：如果以前的值是偶数奇偶校验数，但新值不是偶数奇偶校验数
子数组中偶数奇偶校验数的计数减少，因此我们更新数组并将-1添加到每个范围。要更新的索引i是分段树的一部分 - 情况4：如果先前的值不是偶数奇偶校验数，但是新的值是偶数奇偶校验数
子数组中偶数奇偶校验数的计数增加，因此我们更新了数组并向每个范围加1。要更新的索引i是分段树的一部分
- 情况1：如果先前值和新值都为偶校验数字
下面是上述方法的实现：
C++
// C++ implementation to find
// number of Even Parity numbers
// in a subarray and performing updates
#include
using namespace std;
#define MAX 1000
// Function that returns true if count
// of set bits in x is even
bool isEvenParity(int x)
{
// parity will store the
// count of set bits
int parity = 0;
while (x != 0) {
if (x & 1)
parity++;
x = x >> 1;
}
if (parity % 2 == 0)
return true;
else
return false;
}
// A utility function to get
// the middle index
int getMid(int s, int e)
{
return s + (e - s) / 2;
}
// Recursive function to get the number
// of Even Parity numbers in a given range
int queryEvenParityUtil(
int* segmentTree, int segmentStart,
int segmentEnd, int queryStart,
int queryEnd, int index)
{
// If segment of this node is a part
// of given range, then return
// the number of Even Parity numbers
// in the segment
if (queryStart <= segmentStart
&& queryEnd >= segmentEnd)
return segmentTree[index];
// If segment of this node
// is outside the given range
if (segmentEnd < queryStart
|| segmentStart > queryEnd)
return 0;
// If a part of this segment
// overlaps with the given range
int mid = getMid(segmentStart, segmentEnd);
return queryEvenParityUtil(
segmentTree, segmentStart, mid,
queryStart, queryEnd, 2 * index + 1)
+ queryEvenParityUtil(
segmentTree, mid + 1, segmentEnd,
queryStart, queryEnd, 2 * index + 2);
}
// Recursive function to update
// the nodes which have the given
// index in their range
void updateValueUtil(
int* segmentTree, int segmentStart,
int segmentEnd, int i, int diff, int si)
{
// Base Case:
if (i < segmentStart || i > segmentEnd)
return;
// If the input index is in range
// of this node, then update the value
// of the node and its children
segmentTree[si] = segmentTree[si] + diff;
if (segmentEnd != segmentStart) {
int mid = getMid(
segmentStart, segmentEnd);
updateValueUtil(
segmentTree, segmentStart,
mid, i, diff, 2 * si + 1);
updateValueUtil(
segmentTree, mid + 1, segmentEnd,
i, diff, 2 * si + 2);
}
}
// Function to update a value in the
// input array and segment tree
void updateValue(int arr[], int* segmentTree,
int n, int i, int new_val)
{
// Check for erroneous input index
if (i < 0 || i > n - 1) {
printf("Invalid Input");
return;
}
int diff, oldValue;
oldValue = arr[i];
// Update the value in array
arr[i] = new_val;
// Case 1: Old and new values
// both are Even Parity numbers
if (isEvenParity(oldValue)
&& isEvenParity(new_val))
return;
// Case 2: Old and new values
// both not Even Parity numbers
if (!isEvenParity(oldValue)
&& !isEvenParity(new_val))
return;
// Case 3: Old value was Even Parity,
// new value is non Even Parity
if (isEvenParity(oldValue)
&& !isEvenParity(new_val)) {
diff = -1;
}
// Case 4: Old value was non Even Parity,
// new_val is Even Parity
if (!isEvenParity(oldValue)
&& !isEvenParity(new_val)) {
diff = 1;
}
// Update the values of
// nodes in segment tree
updateValueUtil(segmentTree, 0,
n - 1, i, diff, 0);
}
// Return number of Even Parity numbers
void queryEvenParity(int* segmentTree,
int n, int queryStart,
int queryEnd)
{
int EvenParityInRange
= queryEvenParityUtil(
segmentTree, 0, n - 1,
queryStart, queryEnd, 0);
cout << EvenParityInRange << "\n";
}
// Recursive function that constructs
// Segment Tree for the given array
int constructSTUtil(int arr[],
int segmentStart,
int segmentEnd,
int* segmentTree,
int si)
{
// If there is one element in array,
// check if it is Even Parity number
// then store 1 in the segment tree
// else store 0 and return
if (segmentStart == segmentEnd) {
// if arr[segmentStart] is
// Even Parity number
if (isEvenParity(arr[segmentStart]))
segmentTree[si] = 1;
else
segmentTree[si] = 0;
return segmentTree[si];
}
// If there are more than one elements,
// then recur for left and right subtrees
// and store the sum of the
// two values in this node
int mid = getMid(segmentStart,
segmentEnd);
segmentTree[si]
= constructSTUtil(
arr, segmentStart, mid,
segmentTree, si * 2 + 1)
+ constructSTUtil(
arr, mid + 1, segmentEnd,
segmentTree, si * 2 + 2);
return segmentTree[si];
}
// Function to construct a segment
// tree from given array
int* constructST(int arr[], int n)
{
// Height of segment tree
int x = (int)(ceil(log2(n)));
// Maximum size of segment tree
int max_size = 2 * (int)pow(2, x) - 1;
int* segmentTree = new int[max_size];
// Fill the allocated memory st
constructSTUtil(arr, 0, n - 1,
segmentTree, 0);
// Return the constructed segment tree
return segmentTree;
}
// Driver Code
int main()
{
int arr[] = { 18, 15, 8, 9, 14, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
// Build segment tree from given array
int* segmentTree = constructST(arr, n);
// Query 1: Query(start = 0, end = 4)
int start = 0;
int end = 4;
queryEvenParity(segmentTree, n, start, end);
// Query 2: Update(i = 3, x = 11),
// i.e Update a[i] to x
int i = 3;
int x = 11;
updateValue(arr, segmentTree, n, i, x);
// Query 3: Query(start = 0, end = 4)
start = 0;
end = 4;
queryEvenParity(
segmentTree, n, start, end);
return 0;
} C#
// C# implementation to find
// number of Even Parity numbers
// in a subarray and performing updates
using System;
class GFG{
public static int MAX = 1000;
// Function that returns true if count
// of set bits in x is even
static bool isEvenParity(int x)
{
// parity will store the
// count of set bits
int parity = 0;
while (x != 0)
{
if ((x & 1) != 0)
parity++;
x = x >> 1;
}
if (parity % 2 == 0)
return true;
else
return false;
}
// A utility function to get
// the middle index
static int getMid(int s, int e)
{
return s + (e - s) / 2;
}
// Recursive function to get the number
// of Even Parity numbers in a given range
static int queryEvenParityUtil(int[] segmentTree,
int segmentStart,
int segmentEnd,
int queryStart,
int queryEnd, int index)
{
// If segment of this node is a part
// of given range, then return
// the number of Even Parity numbers
// in the segment
if (queryStart <= segmentStart &&
queryEnd >= segmentEnd)
return segmentTree[index];
// If segment of this node
// is outside the given range
if (segmentEnd < queryStart ||
segmentStart > queryEnd)
return 0;
// If a part of this segment
// overlaps with the given range
int mid = getMid(segmentStart, segmentEnd);
return queryEvenParityUtil(segmentTree, segmentStart,
mid, queryStart, queryEnd,
2 * index + 1) +
queryEvenParityUtil(segmentTree, mid + 1,
segmentEnd, queryStart,
queryEnd, 2 * index + 2);
}
// Recursive function to update
// the nodes which have the given
// index in their range
static void updateValueUtil(int[] segmentTree,
int segmentStart,
int segmentEnd, int i,
int diff, int si)
{
// Base Case:
if (i < segmentStart || i > segmentEnd)
return;
// If the input index is in range
// of this node, then update the value
// of the node and its children
segmentTree[si] = segmentTree[si] + diff;
if (segmentEnd != segmentStart)
{
int mid = getMid(segmentStart, segmentEnd);
updateValueUtil(segmentTree, segmentStart, mid,
i, diff, 2 * si + 1);
updateValueUtil(segmentTree, mid + 1,
segmentEnd, i, diff,
2 * si + 2);
}
}
// Function to update a value in the
// input array and segment tree
static void updateValue(int[] arr, int[] segmentTree,
int n, int i, int new_val)
{
// Check for erroneous input index
if (i < 0 || i > n - 1)
{
Console.WriteLine("Invalid Input");
return;
}
int diff = 0, oldValue = 0;
oldValue = arr[i];
// Update the value in array
arr[i] = new_val;
// Case 1: Old and new values
// both are Even Parity numbers
if (isEvenParity(oldValue) &&
isEvenParity(new_val))
return;
// Case 2: Old and new values
// both not Even Parity numbers
if (!isEvenParity(oldValue) &&
!isEvenParity(new_val))
return;
// Case 3: Old value was Even Parity,
// new value is non Even Parity
if (isEvenParity(oldValue) &&
!isEvenParity(new_val))
{
diff = -1;
}
// Case 4: Old value was non Even Parity,
// new_val is Even Parity
if (!isEvenParity(oldValue) &&
!isEvenParity(new_val))
{
diff = 1;
}
// Update the values of
// nodes in segment tree
updateValueUtil(segmentTree, 0, n - 1, i, diff, 0);
}
// Return number of Even Parity numbers
static void queryEvenParity(int[] segmentTree, int n,
int queryStart,
int queryEnd)
{
int EvenParityInRange = queryEvenParityUtil(
segmentTree, 0, n - 1, queryStart, queryEnd, 0);
Console.WriteLine(EvenParityInRange);
}
// Recursive function that constructs
// Segment Tree for the given array
static int constructSTUtil(int[] arr, int segmentStart,
int segmentEnd,
int[] segmentTree, int si)
{
// If there is one element in array,
// check if it is Even Parity number
// then store 1 in the segment tree
// else store 0 and return
if (segmentStart == segmentEnd)
{
// if arr[segmentStart] is
// Even Parity number
if (isEvenParity(arr[segmentStart]))
segmentTree[si] = 1;
else
segmentTree[si] = 0;
return segmentTree[si];
}
// If there are more than one elements,
// then recur for left and right subtrees
// and store the sum of the
// two values in this node
int mid = getMid(segmentStart, segmentEnd);
segmentTree[si] = constructSTUtil(arr, segmentStart, mid,
segmentTree, si * 2 + 1) +
constructSTUtil(arr, mid + 1, segmentEnd,
segmentTree, si * 2 + 2);
return segmentTree[si];
}
// Function to construct a segment
// tree from given array
static int[] constructST(int[] arr, int n)
{
// Height of segment tree
int x = (int)(Math.Ceiling(Math.Log(n, 2)));
// Maximum size of segment tree
int max_size = 2 * (int)Math.Pow(2, x) - 1;
int[] segmentTree = new int[max_size];
// Fill the allocated memory st
constructSTUtil(arr, 0, n - 1, segmentTree, 0);
// Return the constructed segment tree
return segmentTree;
}
// Driver Code
public static void Main()
{
int[] arr = { 18, 15, 8, 9, 14, 5 };
int n = arr.Length;
// Build segment tree from given array
int[] segmentTree = constructST(arr, n);
// Query 1: Query(start = 0, end = 4)
int start = 0;
int end = 4;
queryEvenParity(segmentTree, n, start, end);
// Query 2: Update(i = 3, x = 11),
// i.e Update a[i] to x
int i = 3;
int x = 11;
updateValue(arr, segmentTree, n, i, x);
// Query 3: Query(start = 0, end = 4)
start = 0;
end = 4;
queryEvenParity(segmentTree, n, start, end);
}
}
// This code is contributed by ukasp输出：
3
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