给定N个元素的数组arr [] ，任务是回答Q个查询，每个查询都有两个整数L和R。对于每个查询，任务是找到数字总和为偶数的子数组arr [L…R]中的元素数量。
例子：
Input: arr[] = {7, 3, 19, 13, 5, 4}
query = { 1, 5 }
Output: 3
Explanation:
Elements 19, 13 and 4 have even digit sum
in the subarray {3, 9, 13, 5, 4}.
Input: arr[] = {0, 1, 2, 3, 4, 5, 6, 7}
query = { 3, 5 }
Output: 1
Explanation:
Only 4 has even digit sum
in the subarray {3, 4, 5}.
天真的方法：
- 只需从索引L到R遍历数组，即可找到每个查询的答案，并且只要数组元素具有偶数位总和，就将计数加1 。这种方法的时间复杂度将是O(n * q) 。
高效的方法：
这个想法是建立一个细分树。
- 段树的表示形式：
- 叶节点是输入数组的元素。
- 每个内部节点包含的叶子数在其下具有所有叶子的偶数和。
- 从给定数组构造细分树：
- 我们从一个段arr [0开始。 。 。 n-1]。每次将当前段分为两半(如果尚未将其变成长度为1的段)，然后在两个半段上调用相同的过程，则对于每个这样的段，我们存储具有偶数和的元素数它下面的所有节点。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the digit sum
// for a number
int digitSum(int num)
{
int sum = 0;
while (num) {
sum += (num % 10);
num /= 10;
}
return sum;
}
// Procedure to build the segment tree
void buildTree(vector& tree, int* arr,
int index, int s, int e)
{
// Reached the leaf node
// of the segment tree
if (s == e) {
if (digitSum(arr[s]) & 1)
tree[index] = 0;
else
tree[index] = 1;
return;
}
// Recursively call the buildTree
// on both the nodes of the tree
int mid = (s + e) / 2;
buildTree(tree, arr, 2 * index,
s, mid);
buildTree(tree, arr, 2 * index + 1,
mid + 1, e);
tree[index] = tree[2 * index]
+ tree[2 * index + 1];
}
// Query procedure to get the answer
// for each query l and r are
// query range
int query(vector tree, int index,
int s, int e, int l, int r)
{
// Out of bound or no overlap
if (r < s || l > e)
return 0;
// Complete overlap
// Query range completely lies in
// the segment tree node range
if (s >= l && e <= r) {
return tree[index];
}
// Partially overlap
// Query range partially lies in
// the segment tree node range
int mid = (s + e) / 2;
return (query(tree, 2 * index, s,
mid, l, r)
+ query(tree, 2 * index + 1,
mid + 1, e, l, r));
}
// Driver code
int main()
{
int arr[] = { 7, 3, 19, 13, 5, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
vector tree(4 * n + 1);
int L = 1, R = 5;
buildTree(tree, arr, 1, 0, n - 1);
cout << query(tree, 1, 0, n - 1, L, R)
<< endl;
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG{
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Procedure to build the segment tree
static void buildTree(int []tree, int []arr,
int index, int s, int e)
{
// Reached the leaf node
// of the segment tree
if (s == e)
{
if (digitSum(arr[s]) % 2 == 1)
tree[index] = 0;
else
tree[index] = 1;
return;
}
// Recursively call the buildTree
// on both the nodes of the tree
int mid = (s + e) / 2;
buildTree(tree, arr, 2 * index,
s, mid);
buildTree(tree, arr, 2 * index + 1,
mid + 1, e);
tree[index] = tree[2 * index] +
tree[2 * index + 1];
}
// Query procedure to get the answer
// for each query l and r are
// query range
static int query(int []tree, int index,
int s, int e,
int l, int r)
{
// Out of bound or no overlap
if (r < s || l > e)
return 0;
// Complete overlap
// Query range completely lies in
// the segment tree node range
if (s >= l && e <= r)
{
return tree[index];
}
// Partially overlap
// Query range partially lies in
// the segment tree node range
int mid = (s + e) / 2;
return (query(tree, 2 * index, s,
mid, l, r) +
query(tree, 2 * index + 1,
mid + 1, e, l, r));
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 7, 3, 19, 13, 5, 4 };
int n = arr.length;
int []tree = new int[4 * n + 1];
int L = 1, R = 5;
buildTree(tree, arr, 1, 0, n - 1);
System.out.print(query(tree, 1, 0,
n - 1, L, R) + "\n");
}
}
// This code is contributed by gauravrajput1Python3
# Python3 implementation of the above approach
# Function to find the digit sum
# for a number
def digitSum(num):
sum = 0;
while (num):
sum += (num % 10)
num //= 10
return sum
# Procedure to build the segment tree
def buildTree(tree, arr, index, s, e):
# Reached the leaf node
# of the segment tree
if (s == e):
if (digitSum(arr[s]) & 1):
tree[index] = 0
else:
tree[index] = 1
return
# Recursively call the buildTree
# on both the nodes of the tree
mid = (s + e) // 2
buildTree(tree, arr, 2 * index,
s, mid)
buildTree(tree, arr, 2 * index + 1,
mid + 1, e)
tree[index] = (tree[2 * index] +
tree[2 * index + 1])
# Query procedure to get the answer
# for each query l and r are
# query range
def query(tree, index, s, e, l, r):
# Out of bound or no overlap
if (r < s or l > e):
return 0
# Complete overlap
# Query range completely lies in
# the segment tree node range
if (s >= l and e <= r):
return tree[index]
# Partially overlap
# Query range partially lies in
# the segment tree node range
mid = (s + e) // 2
return (query(tree, 2 * index,
s, mid, l, r) +
query(tree, 2 * index + 1,
mid + 1, e, l, r))
# Driver code
arr = [ 7, 3, 19, 13, 5, 4 ]
n = len(arr)
tree = [0] * (4 * n + 1)
L = 1
R = 5
buildTree(tree, arr, 1, 0, n - 1);
print(query(tree, 1, 0, n - 1, L, R))
# This code is contributed by ApurvarajC#
// C# implementation of the approach
using System;
class GFG{
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Procedure to build the segment tree
static void buildTree(int []tree, int []arr,
int index, int s, int e)
{
// Reached the leaf node
// of the segment tree
if (s == e)
{
if (digitSum(arr[s]) % 2 == 1)
tree[index] = 0;
else
tree[index] = 1;
return;
}
// Recursively call the buildTree
// on both the nodes of the tree
int mid = (s + e) / 2;
buildTree(tree, arr, 2 * index,
s, mid);
buildTree(tree, arr, 2 * index + 1,
mid + 1, e);
tree[index] = tree[2 * index] +
tree[2 * index + 1];
}
// Query procedure to get the answer
// for each query l and r are
// query range
static int query(int []tree, int index,
int s, int e,
int l, int r)
{
// Out of bound or no overlap
if (r < s || l > e)
return 0;
// Complete overlap
// Query range completely lies in
// the segment tree node range
if (s >= l && e <= r)
{
return tree[index];
}
// Partially overlap
// Query range partially lies in
// the segment tree node range
int mid = (s + e) / 2;
return (query(tree, 2 * index, s,
mid, l, r) +
query(tree, 2 * index + 1,
mid + 1, e, l, r));
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 7, 3, 19, 13, 5, 4 };
int n = arr.Length;
int []tree = new int[4 * n + 1];
int L = 1, R = 5;
buildTree(tree, arr, 1, 0, n - 1);
Console.Write(query(tree, 1, 0,
n - 1, L, R) + "\n");
}
}
// This code is contributed by gauravrajput1输出：
3
时间复杂度： O(Q * log(N))