给定一个由N个整数组成的数组Arr和Q个查询，每个查询的范围从L到R。查找范围从L到R的具有最大位数和的元素，如果一个以上的元素具有最大位数的和，则从中找出最大元素。
例子：
Input: Arr[] = { 16, 12, 43, 55}
Q = 2
L = 0, R = 3
L = 0, R = 2
Output: 55
43
Explanation:
The range (0, 3) in the 1st query has
[16, 12, 43, 55]. Here, the digit sums are:
for 16, 1 + 6 = 7
for 12, 1 + 2 = 3
for 43, 4 + 3 = 7
for 55, 5 + 5 = 10
Hence, the max digit sum is 10 and
max digit sum value is 55.
The range (0, 2) in the 1st query has
[16, 12, 43]. Here, the digit sums are:
for 16, 1 + 6 = 7
for 12, 1 + 2 = 3
for 43, 4 + 3 = 7
Hence, the max digit sum is 7 and
max digit sum value is max( 16, 43) = 43.
天真的方法：
一个简单的解决方案是运行一个从L到R的循环，计算每个元素的数字总和，并为每个查询找到从L到R的最大数字总和元素。
时间复杂度：O(Q * N)，
辅助空间：O(1)
高效方法：
- 一种有效的方法是构建一个“分段树”，其中每个节点都存储两个值( value和max_digit_sum )，并对其进行范围查询以找到最大位数和和相应的元素。但是对于构建段树，我们必须考虑在树的节点上存储什么。
- 为了找出最大数字总和值，我们需要做两件事，一个是数值，另一个是数字总和。合并将返回两件事，数字总和将在段树中存储max(max_digit_sum.left，max_digit_sum.right)，而value将包含相应的元素值。
- 如果我们深入研究，则任意两个范围组合的最大位数总和要么是左侧的最大位数总和，要么是右侧的最大位数总和，以最大值为准。
- 段树的表示形式：
- 叶节点是给定数组的元素。
- 每个内部节点代表叶节点的一些合并。对于不同的问题，合并可能会有所不同。对于此问题，合并是节点下叶子的max_digit_sum的最大值。
- 树的数组表示形式用于表示段树。对于索引i处的每个节点，左子节点在索引2 * i + 1处，右子节点在索引2 * i + 2处，父节点在(i-1)/ 2处。
- 从给定数组构造细分树：
- 我们从一个段arr [0开始。 。 。 n-1]。每次我们将当前段分为两半(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，我们将max_digit_sum和值存储在对应的节点。
- 然后，我们在细分树上进行范围查询，以找出给定范围的max_digit_sum并输出相应的值。
下面是上述方法的实现。
C++
// C++ program to find
// maximum digit sum value
#include
using namespace std;
// Struct two store two values in one node
struct Node {
int value;
int max_digit_sum;
};
Node tree[4 * 10000];
// Function to find the digit sum
// for a number
int digitSum(int x)
{
int sum = 0;
while (x) {
sum += (x % 10);
x /= 10;
}
}
// Function to build the segment tree
void build(int a[], int index, int beg, int end)
{
if (beg == end) {
// If there is one element in array,
tree[index].value = a[beg];
tree[index].max_digit_sum = digitSum(a[beg]);
}
else {
int mid = (beg + end) / 2;
// If there are more than one elements,
// then recur for left and right subtrees
build(a, 2 * index + 1, beg, mid);
build(a, 2 * index + 2, mid + 1, end);
if (tree[2 * index + 1].max_digit_sum >
tree[2 * index + 2].max_digit_sum)
{
tree[index].max_digit_sum =
tree[2 * index + 1].max_digit_sum;
tree[index].value =
tree[2 * index + 1].value;
}
else if (tree[2 * index + 2].max_digit_sum >
tree[2 * index + 1].max_digit_sum)
{
tree[index].max_digit_sum =
tree[2 * index + 2].max_digit_sum;
tree[index].value =
tree[2 * index + 2].value;
}
else
{
tree[index].max_digit_sum =
tree[2 * index + 2].max_digit_sum;
tree[index].value =
max(tree[2 * index + 2].value,
tree[2 * index + 1].value);
}
}
}
// Function to do the range query in the segment
// tree for the maximum digit sum
Node query(int index, int beg, int end,
int l, int r)
{
Node result;
result.value = result.max_digit_sum = -1;
// If segment of this node is outside the given
// range, then return the minimum valueue.
if (beg > r || end < l)
return result;
// If segment of this node is a part of given
// range, then return the node of the segment
if (beg >= l && end <= r)
return tree[index];
int mid = (beg + end) / 2;
// If left segment of this node falls out of
// range, then recur in the right side of
// the tree
if (l > mid)
return query(2 * index + 2, mid + 1,
end, l, r);
// If right segment of this node falls out of
// range, then recur in the left side of
// the tree
if (r <= mid)
return query(2 * index + 1, beg,
mid, l, r);
// If a part of this segment overlaps with
// the given range
Node left = query(2 * index + 1, beg,
mid, l, r);
Node right = query(2 * index + 2, mid + 1,
end, l, r);
if (left.max_digit_sum > right.max_digit_sum)
{
result.max_digit_sum = left.max_digit_sum;
result.value = left.value;
}
else if (right.max_digit_sum > left.max_digit_sum)
{
result.max_digit_sum = right.max_digit_sum;
result.value = right.value;
}
else
{
result.max_digit_sum = left.max_digit_sum;
result.value = max(right.value, left.value);
}
// Returns the value
return result;
}
// Driver code
int main()
{
int a[] = {16, 12, 43, 55};
// Calculates the length of array
int N = sizeof(a) / sizeof(a[0]);
// Calls the build function to build
// the segment tree
build(a, 0, 0, N - 1);
// Find the max digit-sum valueue between
// 0th and 3rd index of array
cout << query(0, 0, N - 1, 0, 3).value
<< endl;
// Find the max digit-sum value between
// 0th and 2nd index of array
cout << query(0, 0, N - 1, 0, 2).value
<< endl;
return 0;
} Java
// Java program to find
// maximum digit sum value
import java.util.*;
class GFG{
// Struct two store two values
// in one node
static class Node
{
int value;
int max_digit_sum;
};
static Node []tree = new Node[4 * 10000];
// Function to find the digit sum
// for a number
static int digitSum(int x)
{
int sum = 0;
while (x > 0)
{
sum += (x % 10);
x /= 10;
}
return sum;
}
// Function to build the segment tree
static void build(int a[], int index,
int beg, int end)
{
if (beg == end)
{
// If there is one element in array,
tree[index].value = a[beg];
tree[index].max_digit_sum = digitSum(a[beg]);
}
else
{
int mid = (beg + end) / 2;
// If there are more than one elements,
// then recur for left and right subtrees
build(a, 2 * index + 1, beg, mid);
build(a, 2 * index + 2, mid + 1, end);
if (tree[2 * index + 1].max_digit_sum >
tree[2 * index + 2].max_digit_sum)
{
tree[index].max_digit_sum =
tree[2 * index + 1].max_digit_sum;
tree[index].value =
tree[2 * index + 1].value;
}
else if (tree[2 * index + 2].max_digit_sum >
tree[2 * index + 1].max_digit_sum)
{
tree[index].max_digit_sum =
tree[2 * index + 2].max_digit_sum;
tree[index].value =
tree[2 * index + 2].value;
}
else
{
tree[index].max_digit_sum =
tree[2 * index + 2].max_digit_sum;
tree[index].value =
Math.max(tree[2 * index + 2].value,
tree[2 * index + 1].value);
}
}
}
// Function to do the range query in the segment
// tree for the maximum digit sum
static Node query(int index, int beg, int end,
int l, int r)
{
Node result = new Node();
result.value = result.max_digit_sum = -1;
// If segment of this node is outside the given
// range, then return the minimum valueue.
if (beg > r || end < l)
return result;
// If segment of this node is a part of given
// range, then return the node of the segment
if (beg >= l && end <= r)
return tree[index];
int mid = (beg + end) / 2;
// If left segment of this node falls out of
// range, then recur in the right side of
// the tree
if (l > mid)
return query(2 * index + 2, mid + 1,
end, l, r);
// If right segment of this node falls out of
// range, then recur in the left side of
// the tree
if (r <= mid)
return query(2 * index + 1, beg,
mid, l, r);
// If a part of this segment overlaps with
// the given range
Node left = query(2 * index + 1, beg,
mid, l, r);
Node right = query(2 * index + 2, mid + 1,
end, l, r);
if (left.max_digit_sum > right.max_digit_sum)
{
result.max_digit_sum = left.max_digit_sum;
result.value = left.value;
}
else if (right.max_digit_sum > left.max_digit_sum)
{
result.max_digit_sum = right.max_digit_sum;
result.value = right.value;
}
else
{
result.max_digit_sum = left.max_digit_sum;
result.value = Math.max(right.value,
left.value);
}
// Returns the value
return result;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 16, 12, 43, 55 };
// Calculates the length of array
int N = a.length;
for(int i = 0; i < tree.length; i++)
tree[i] = new Node();
// Calls the build function to build
// the segment tree
build(a, 0, 0, N - 1);
// Find the max digit-sum valueue between
// 0th and 3rd index of array
System.out.print(
query(0, 0, N - 1, 0, 3).value + "\n");
// Find the max digit-sum value between
// 0th and 2nd index of array
System.out.print(
query(0, 0, N - 1, 0, 2).value + "\n");
}
}
// This code is contributed by Amit KatiyarPython3
# Python3 program to find
# maximum digit sum value
# Struct two store two values in one node
class Node:
def __init__(self):
self.value = 0
self.max_digit_sum = 0
tree = [Node() for i in range(4 * 10000)]
# Function to find the digit sum
# for a number
def digitSum(x):
sum = 0;
while(x != 0):
sum += (x % 10)
x //= 10
return sum
# Function to build the segment tree
def build(a, index, beg, end):
if (beg == end):
# If there is one element in array,
tree[index].value = a[beg]
tree[index].max_digit_sum = digitSum(a[beg])
else:
mid = (beg + end) // 2
# If there are more than one elements,
# then recur for left and right subtrees
build(a, 2 * index + 1, beg, mid)
build(a, 2 * index + 2, mid + 1, end)
if (tree[2 * index + 1].max_digit_sum >
tree[2 * index + 2].max_digit_sum):
tree[index].max_digit_sum = tree[2 * index + 1].max_digit_sum
tree[index].value = tree[2 * index + 1].value
elif (tree[2 * index + 2].max_digit_sum >
tree[2 * index + 1].max_digit_sum):
tree[index].max_digit_sum = tree[2 * index + 2].max_digit_sum
tree[index].value = tree[2 * index + 2].value
else:
tree[index].max_digit_sum = tree[2 * index + 2].max_digit_sum
tree[index].value = max(tree[2 * index + 2].value,
tree[2 * index + 1].value)
# Function to do the range query in the segment
# tree for the maximum digit sum
def query(index, beg, end, l, r):
result = Node()
result.value = result.max_digit_sum = -1
# If segment of this node is outside the given
# range, then return the minimum valueue.
if (beg > r or end < l):
return result
# If segment of this node is a part of given
# range, then return the node of the segment
if (beg >= l and end <= r):
return tree[index]
mid = (beg + end) // 2
# If left segment of this node falls out of
# range, then recur in the right side of
# the tree
if (l > mid):
return query(2 * index + 2, mid + 1, end, l, r)
# If right segment of this node falls out of
# range, then recur in the left side of
# the tree
if (r <= mid):
return query(2 * index + 1, beg, mid, l, r)
# If a part of this segment overlaps with
# the given range
left = query(2 * index + 1, beg, mid, l, r)
right = query(2 * index + 2, mid + 1, end, l, r)
if (left.max_digit_sum > right.max_digit_sum):
result.max_digit_sum = left.max_digit_sum
result.value = left.value
elif (right.max_digit_sum > left.max_digit_sum):
result.max_digit_sum = right.max_digit_sum
result.value = right.value
else:
result.max_digit_sum = left.max_digit_sum
result.value = max(right.value, left.value)
# Returns the value
return result
# Driver code
if __name__=="__main__":
a = [ 16, 12, 43, 55 ]
# Calculates the length of array
N = len(a)
# Calls the build function to build
# the segment tree
build(a, 0, 0, N - 1)
# Find the max digit-sum valueue between
# 0th and 3rd index of array
print(query(0, 0, N - 1, 0, 3).value)
# Find the max digit-sum value between
# 0th and 2nd index of array
print(query(0, 0, N - 1, 0, 2).value)
# This code is contributed by rutvik_56C#
// C# program to find
// maximum digit sum value
using System;
class GFG{
// Struct two store
// two values in one node
class Node
{
public int value;
public int max_digit_sum;
};
static Node []tree = new Node[4 * 10000];
// Function to find the digit sum
// for a number
static int digitSum(int x)
{
int sum = 0;
while (x > 0)
{
sum += (x % 10);
x /= 10;
}
return sum;
}
// Function to build the segment tree
static void build(int []a, int index,
int beg, int end)
{
if (beg == end)
{
// If there is one element in array,
tree[index].value = a[beg];
tree[index].max_digit_sum =
digitSum(a[beg]);
}
else
{
int mid = (beg + end) / 2;
// If there are more than one elements,
// then recur for left and right subtrees
build(a, 2 * index + 1, beg, mid);
build(a, 2 * index + 2, mid + 1, end);
if (tree[2 * index + 1].max_digit_sum >
tree[2 * index + 2].max_digit_sum)
{
tree[index].max_digit_sum =
tree[2 * index + 1].max_digit_sum;
tree[index].value =
tree[2 * index + 1].value;
}
else if (tree[2 * index + 2].max_digit_sum >
tree[2 * index + 1].max_digit_sum)
{
tree[index].max_digit_sum =
tree[2 * index + 2].max_digit_sum;
tree[index].value =
tree[2 * index + 2].value;
}
else
{
tree[index].max_digit_sum =
tree[2 * index + 2].max_digit_sum;
tree[index].value =
Math.Max(tree[2 * index + 2].value,
tree[2 * index + 1].value);
}
}
}
// Function to do the range query
// in the segment tree for the
// maximum digit sum
static Node query(int index, int beg,
int end, int l, int r)
{
Node result = new Node();
result.value = result.max_digit_sum = -1;
// If segment of this node is
// outside the given range,
// then return the minimum valueue.
if (beg > r || end < l)
return result;
// If segment of this node
// is a part of given range,
// then return the node of the segment
if (beg >= l && end <= r)
return tree[index];
int mid = (beg + end) / 2;
// If left segment of this
// node falls out of range,
// then recur in the right
// side of the tree
if (l > mid)
return query(2 * index + 2, mid + 1,
end, l, r);
// If right segment of this
// node falls out of range,
// then recur in the left side of
// the tree
if (r <= mid)
return query(2 * index + 1, beg,
mid, l, r);
// If a part of this segment
// overlaps with the given range
Node left = query(2 * index + 1, beg,
mid, l, r);
Node right = query(2 * index + 2, mid + 1,
end, l, r);
if (left.max_digit_sum > right.max_digit_sum)
{
result.max_digit_sum = left.max_digit_sum;
result.value = left.value;
}
else if (right.max_digit_sum > left.max_digit_sum)
{
result.max_digit_sum = right.max_digit_sum;
result.value = right.value;
}
else
{
result.max_digit_sum = left.max_digit_sum;
result.value = Math.Max(right.value,
left.value);
}
// Returns the value
return result;
}
// Driver code
public static void Main(String[] args)
{
int []a = {16, 12, 43, 55};
// Calculates the length
// of array
int N = a.Length;
for(int i = 0; i < tree.Length; i++)
tree[i] = new Node();
// Calls the build function
// to build the segment tree
build(a, 0, 0, N - 1);
// Find the max digit-sum value between
// 0th and 3rd index of array
Console.Write(query(0, 0,
N - 1,
0, 3).value + "\n");
// Find the max digit-sum value between
// 0th and 2nd index of array
Console.Write(query(0, 0,
N - 1,
0, 2).value + "\n");
}
}
// This code is contributed by Rajput-Ji输出：
55
43
复杂度分析：
树木构建的时间复杂度为O(N)。总共有2n-1个节点，并且在树结构中每个节点的值仅计算一次。
每个查询的时间复杂度为O(log N)。
该问题的时间复杂度为O(Q * log N)
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