在[L，R]范围内最多K个移动中最大化数字的总和

给定一个由N个整数和Q个查询组成的数组arr [] 。每个查询由3个整数L ， R和K组成。您可以单步从索引i移至索引i + 1 ，也可以单步停留在该特定索引中。您可以从L的最大的k步移动到右食指和打印每次你在每一个步骤，包括第L号数的总和。任务是在最多K个动作中使总和最大化。如果我们不能以K步从L移到R ，则打印“ No” 。

例子：

Input: arr[] = {1, 3, 2, -4, -5}, q = {
{0, 2, 2},
{0, 2, 4},
{3, 4, 1},
{0, 4, 2}}
Output:
6
12
-9
No

For first query:
In first step move from 0th index to 1st index, hence 1 + 3 = 4
In second step move from 1st index to 2nd index, hence 4 + 2 = 6

For second query:
In first step move from 0th index to 1st index, hence 1 + 3 = 4
In second step stay at the 1st index, hence 4 + 3 = 7
In third step again stay at the 1st index, hence 7 + 3 = 10
In fourth step move from 1st index to 2nd index only, hence 10 + 2 = 12

一个幼稚的方法是首先检查L – R> K ，如果不是，则我们不能以K为步长从L移到R索引。从L迭代到R ，得到L和R之间所有元素的总和。然后找到L和R范围内的最大元素，答案将是(K –(R – L))* max范围内的元素之和。如果最大值为负数，我们将精确执行R – L动作，否则我们将在[L，R]范围内的最大数字索引处执行额外的步骤。
时间复杂度：每个查询为O(R – L)。

一种有效的方法是使用[L，R]范围内的分段树来找到最大数量，并使用前缀和来找到该范围内的和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to create the tree
void tree(int low, int high, int pos,
          int b[], int a[], int n)
{
    // Leaf nodes
    if (low == high) {
        b[pos] = a[high];
        return;
    }
    int mid = (high + low) / 2;
  
    // Left subtree
    tree(low, mid, 2 * pos + 1, b, a, n);
  
    // Right subtree
    tree(mid + 1, high, 2 * pos + 2, b, a, n);
  
    // Merge the maximum
    b[pos] = max(b[2 * pos + 1], b[2 * pos + 2]);
}
  
// Function that returns the maximum in range L and R
int rangemax(int s, int e, int low, int high,
             int pos, int b[], int a[], int n)
{
    // Complete overlap
    if (low <= s && high >= e)
        return b[pos];
  
    // Out of range completely
    if (e < low || s > high)
        return INT_MIN;
    int mid = (s + e) / 2;
  
    // Find maximum in left and right subtrees
    int left = rangemax(s, mid, low, high,
                        2 * pos + 1, b, a, n);
    int right = rangemax(mid + 1, e, low, high,
                         2 * pos + 2, b, a, n);
  
    // Return the maximum of both
    return max(left, right);
}
  
// Function that solves a query
int solveQuery(int l, int r, int k, int n, int a[],
               int b[], int prefix[])
{
  
    // If there are ko
    if (r - l > k)
        return -1;
  
    // Find maximum in range L and R
    int maximum = rangemax(0, n - 1, l, r, 0, b, a, n);
  
    // If maximum is 0
    if (maximum < 0)
        maximum = 0;
  
    // Find the prefix sum
    int rangesum = prefix[r];
  
    // If not first element
    if (l > 0)
        rangesum -= prefix[l - 1];
  
    // Get the answer
    int answer = rangesum + (k - (r - l)) * maximum;
  
    return answer;
}
  
// Function that solves the queries
void solveQueries(int n, int a[], int b[],
                  int prefix[], int queries[][3], int q)
{
  
    // Solve all the queries
    for (int i = 0; i < q; i++) {
        int ans = solveQuery(queries[i][0], queries[i][1],
                             queries[i][2], n, a, b, prefix);
        if (ans != -1)
            cout << ans << endl;
        else
            cout << "No" << endl;
    }
}
  
// Function to find the prefix sum
void findPrefixSum(int prefix[], int a[], int n)
{
    prefix[0] = a[0];
    for (int i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }
}
  
// Driver code
int main()
{
    int a[] = { 1, 3, 2, -4, -5 };
    int n = sizeof(a) / sizeof(a[0]);
  
    // Array for segment tree
    int b[5 * n];
  
    // Create segment tree
    tree(0, n - 1, 0, b, a, n);
  
    int prefix[n];
  
    // Fill prefix sum array
    findPrefixSum(prefix, a, n);
  
    // Queries
    int queries[][3] = { { 0, 2, 2 },
                         { 0, 2, 4 },
                         { 3, 4, 1 },
                         { 0, 4, 2 } };
  
    int q = sizeof(queries) / sizeof(queries[0]);
    solveQueries(n, a, b, prefix, queries, q);
  
    return 0;
}

Java

// Java implementation of the approach
  
class GFG
{
  
// Function to create the tree
static void tree(int low, int high, int pos,
        int b[], int a[], int n)
{
    // Leaf nodes
    if (low == high)
    {
        b[pos] = a[high];
        return;
    }
    int mid = (high + low) / 2;
  
    // Left subtree
    tree(low, mid, 2 * pos + 1, b, a, n);
  
    // Right subtree
    tree(mid + 1, high, 2 * pos + 2, b, a, n);
  
    // Merge the maximum
    b[pos] = Math.max(b[2 * pos + 1], b[2 * pos + 2]);
}
  
// Function that returns the maximum in range L and R
static int rangemax(int s, int e, int low, int high,
            int pos, int b[], int a[], int n)
{
    // Complete overlap
    if (low <= s && high >= e)
        return b[pos];
  
    // Out of range completely
    if (e < low || s > high)
        return Integer.MIN_VALUE;
    int mid = (s + e) / 2;
  
    // Find maximum in left and right subtrees
    int left = rangemax(s, mid, low, high,
                        2 * pos + 1, b, a, n);
    int right = rangemax(mid + 1, e, low, high,
                        2 * pos + 2, b, a, n);
  
    // Return the maximum of both
    return Math.max(left, right);
}
  
// Function that solves a query
static int solveQuery(int l, int r, int k, int n, int a[],
                                    int b[], int prefix[])
{
  
    // If there are ko
    if (r - l > k)
        return -1;
  
    // Find maximum in range L and R
    int maximum = rangemax(0, n - 1, l, r, 0, b, a, n);
  
    // If maximum is 0
    if (maximum < 0)
        maximum = 0;
  
    // Find the prefix sum
    int rangesum = prefix[r];
  
    // If not first element
    if (l > 0)
        rangesum -= prefix[l - 1];
  
    // Get the answer
    int answer = rangesum + (k - (r - l)) * maximum;
  
    return answer;
}
  
// Function that solves the queries
static void solveQueries(int n, int a[], int b[],
                int prefix[], int queries[][], int q)
{
  
    // Solve all the queries
    for (int i = 0; i < q; i++)
    {
        int ans = solveQuery(queries[i][0], queries[i][1],
                            queries[i][2], n, a, b, prefix);
        if (ans != -1)
            System.out.println(ans);
        else
            System.out.println("No" );
    }
}
  
// Function to find the prefix sum
static void findPrefixSum(int prefix[], int a[], int n)
{
    prefix[0] = a[0];
    for (int i = 1; i < n; i++) 
    {
        prefix[i] = prefix[i - 1] + a[i];
    }
}
  
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 3, 2, -4, -5 };
    int n = a.length;
  
    // Array for segment tree
    int b[] = new int[5 * n];
  
    // Create segment tree
    tree(0, n - 1, 0, b, a, n);
  
    int prefix[] = new int[n];
  
    // Fill prefix sum array
    findPrefixSum(prefix, a, n);
  
    // Queries
    int queries[][] = { { 0, 2, 2 },
                        { 0, 2, 4 },
                        { 3, 4, 1 },
                        { 0, 4, 2 } };
  
    int q = queries.length;
    solveQueries(n, a, b, prefix, queries, q);
  
}
}
  
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
  
# Function to create the tree
def tree( low, high, pos, b, a, n):
      
    # Leaf nodes
    if (low == high):
        b[pos] = a[high]
        return
      
    mid = (high + low) // 2
      
    # Left subtree
    tree(low, mid, 2 * pos + 1, b, a, n)
      
    # Right subtree
    tree(mid + 1, high, 2 * pos + 2, b, a, n)
      
    # Merge the maximum
    b[pos] = max(b[2 * pos + 1], b[2 * pos + 2])
  
# Function that returns the maximum in range L and R
def rangemax(s, e, low, high, pos, b, a, n):
      
    # Complete overlap
    if (low <= s and high >= e):
        return b[pos]
          
    # Out of range completely
    if (e < low or s > high):
        return -(2**32)
      
    mid = (s + e) // 2
      
    # Find maximum in left and right subtrees
    left = rangemax(s, mid, low, high, 2 * pos + 1, b, a, n)
    right = rangemax(mid + 1, e, low, high, 2 * pos + 2, b, a, n)
      
    # Return the maximum of both
    return max(left, right)
  
# Function that solves a query
def solveQuery(l, r, k, n, a, b, prefix):
      
    # If there are ko
    if (r - l > k):
        return -1
      
    # Find maximum in range L and R
    maximum = rangemax(0, n - 1, l, r, 0, b, a, n)
      
    # If maximum is 0
    if (maximum < 0):
        maximum = 0
      
    # Find the prefix sum
    rangesum = prefix[r]
      
    # If not first element
    if (l > 0):
        rangesum -= prefix[l - 1]
          
    # Get the answer
    answer = rangesum + (k - (r - l)) * maximum
    return answer
  
# Function that solves the queries
def solveQueries( n, a, b, prefix, queries, q):
      
    # Solve all the queries
    for i in range(q):
        ans = solveQuery(queries[i][0], queries[i][1], 
                        queries[i][2], n, a, b, prefix)
        if (ans != -1):
            print(ans)
        else:
            print("No")
      
# Function to find the prefix sum
def findPrefixSum( prefix, a, n):
    prefix[0] = a[0]
    for i in range(1, n):
        prefix[i] = prefix[i - 1] + a[i]
  
# Driver code
a = [1, 3, 2, -4, -5 ]
n = len(a)
  
# Array for segment tree
b = [0]*(5 * n)
  
# Create segment tree
tree(0, n - 1, 0, b, a, n)
  
prefix = [0]*n
  
# Fill prefix sum array
findPrefixSum(prefix, a, n)
  
# Queries
queries= [[0, 2, 2],[0, 2, 4],[3, 4, 1],[0, 4, 2]]
  
q = len(queries)
solveQueries(n, a, b, prefix, queries, q)
  
# This code is contributed by SHUBHAMSINGH10

C#

// C# program to implement
// the above approach
using System;
  
class GFG
{
  
// Function to create the tree
static void tree(int low, int high, int pos,
                    int []b, int []a, int n)
{
    // Leaf nodes
    if (low == high)
    {
        b[pos] = a[high];
        return;
    }
    int mid = (high + low) / 2;
  
    // Left subtree
    tree(low, mid, 2 * pos + 1, b, a, n);
  
    // Right subtree
    tree(mid + 1, high, 2 * pos + 2, b, a, n);
  
    // Merge the maximum
    b[pos] = Math.Max(b[2 * pos + 1], b[2 * pos + 2]);
}
  
// Function that returns the maximum in range L and R
static int rangemax(int s, int e, int low, int high,
            int pos, int []b, int []a, int n)
{
    // Complete overlap
    if (low <= s && high >= e)
        return b[pos];
  
    // Out of range completely
    if (e < low || s > high)
        return int.MinValue;
    int mid = (s + e) / 2;
  
    // Find maximum in left and right subtrees
    int left = rangemax(s, mid, low, high,
                        2 * pos + 1, b, a, n);
    int right = rangemax(mid + 1, e, low, high,
                        2 * pos + 2, b, a, n);
  
    // Return the maximum of both
    return Math.Max(left, right);
}
  
// Function that solves a query
static int solveQuery(int l, int r, int k, int n, int []a,
                                    int []b, int []prefix)
{
  
    // If there are ko
    if (r - l > k)
        return -1;
  
    // Find maximum in range L and R
    int maximum = rangemax(0, n - 1, l, r, 0, b, a, n);
  
    // If maximum is 0
    if (maximum < 0)
        maximum = 0;
  
    // Find the prefix sum
    int rangesum = prefix[r];
  
    // If not first element
    if (l > 0)
        rangesum -= prefix[l - 1];
  
    // Get the answer
    int answer = rangesum + (k - (r - l)) * maximum;
  
    return answer;
}
  
// Function that solves the queries
static void solveQueries(int n, int []a, int []b,
                int []prefix, int [,]queries, int q)
{
  
    // Solve all the queries
    for (int i = 0; i < q; i++)
    {
        int ans = solveQuery(queries[i,0], queries[i,1],
                            queries[i,2], n, a, b, prefix);
        if (ans != -1)
            Console.WriteLine(ans);
        else
            Console.WriteLine("No" );
    }
}
  
// Function to find the prefix sum
static void findPrefixSum(int []prefix, int []a, int n)
{
    prefix[0] = a[0];
    for (int i = 1; i < n; i++) 
    {
        prefix[i] = prefix[i - 1] + a[i];
    }
}
  
// Driver code
public static void Main(String[] args)
{
    int []a = { 1, 3, 2, -4, -5 };
    int n = a.Length;
  
    // Array for segment tree
    int []b = new int[5 * n];
  
    // Create segment tree
    tree(0, n - 1, 0, b, a, n);
  
    int []prefix = new int[n];
  
    // Fill prefix sum array
    findPrefixSum(prefix, a, n);
  
    // Queries
    int [,]queries = { { 0, 2, 2 },
                        { 0, 2, 4 },
                        { 3, 4, 1 },
                        { 0, 4, 2 } };
  
    int q = queries.GetLength(0);
    solveQueries(n, a, b, prefix, queries, q);
  
}
}
  
// This code has been contributed by 29AjayKumar

输出：

6
12
-9
No

时间复杂度：每个查询O(Log N)。
辅助空间： O(N log N)

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