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📜  查询将给定的子数组与给定的数字X相乘并打印总和

📅  最后修改于: 2021-05-17 04:26:16             🧑  作者: Mango

给定数组arr []和Q个查询，其中每个查询包含三个整数(l，r，x)，任务是在将每个元素与x乘以后打印范围[l，r]中的元素之和。 

Query(l, r, x) = 
   arr[l]*x + arr[l+1]*x + .... arr[r-1]*x + arr[r]*x

注意：与x的乘积已完成以计算答案，并且在查询的任何步骤都不会在输入数组中对其进行更新。
例子：

天真的方法：想法是遍历数组的每个查询，并针对每个查询遍历[l，r]范围的元素，并找到每个元素的总和乘以x。
时间复杂度： O(Q * N)
高效方法：想法是预先计算数组的前缀和，然后为每个查询找到范围[l，r]的元素之和，然后乘以x来找到每个查询的答案。
以下是计算每个查询答案的公式： 

// pre_sum[i] denotes the prefix sum of
// the array from the array 0 to i
answer = x * (pre_sum[r] - pre_sum[l-1])

下面是上述方法的实现：

C++
// C++ implementation to multiply
// the given subarray by the x
// for multiple queries of the Q
 
#include 
 
using namespace std;
 
// Function to answer each query
int query(vector pre_sum, int n,
         int l, int r, int val)
{
    int ans;
    int temp = 0;
    if (l > 0) {
        temp = pre_sum[l-1];
    }
    ans = val * (pre_sum[r] - temp);
    return ans;
}
 
// Function to multiply the subarray
// by the x for multiple Queries
void multiplyArray(int arr[],
 vector, int>> q,
                             int n){
    vector pre_sum;
    int s = 0;
     
    // Loop to compute the prefix
    // sum of the array
    for (int i = 0; i < n; i++){
        s += arr[i];
        pre_sum.push_back(s);
    }
     
    // Loop to answer each query
    for(auto i: q){
        cout << query(pre_sum, n, i.first.first,
         i.first.second, i.second) << endl;
    }
}
 
// Driver Code
int main()
{
    // Array
    int arr[] = { 2, 3, 4, 2, 5, 1 };
    int n = 6;
    vector, int>> q;
    q.push_back({{2, 4}, 5});
    q.push_back({{1, 1}, 4});
    q.push_back({{1, 3}, -2});
     
    // Function Call
    multiplyArray(arr, q, n);
     
    return 0;
}

Java
/*package whatever //do not write package name here */
// Java implementation to multiply
// the given subarray by the x
// for multiple queries of the Q
import java.io.*;
import java.util.ArrayList;
class GFG
{
 
  // Function to answer each query
  public static int query(ArrayList pre_sum,
                          int n, int l, int r, int val)
  {
    int ans;
    int temp = 0;
    if (l > 0)
    {
      temp = pre_sum.get(l - 1);
    }
    ans = val * (pre_sum.get(r) - temp);
    return ans;
  }
 
  // Function to multiply the subarray
  // by the x for multiple Queries
  public static void multiplyArray(int arr[],
                                   ArrayList q,
                                   int n)
  {
    ArrayList pre_sum = new ArrayList<>();
    int s = 0;
 
    // Loop to compute the prefix
    // sum of the array
    for (int i = 0; i < n; i++)
    {
      s += arr[i];
      pre_sum.add(s);
    }
 
    // Loop to answer each query
    for(int i = 0; i < q.size(); i++)
    {
      System.out.println(query(pre_sum, n,
                               q.get(i).pr.a,
                               q.get(i).pr.b,
                               q.get(i).second));
    }
  }
 
  // Driver Code
  public static void main(String [] args)
  {
    // Array
    int arr[] = { 2, 3, 4, 2, 5, 1 };
    int n = 6;
    ArrayList q = new ArrayList<>();
    q.add(new pair(new pair1(2, 4), 5));
    q.add(new pair(new pair1(1, 1), 4));
    q.add(new pair(new pair1(1, 3), -2));
 
    // Function Call
    multiplyArray(arr, q, n);
  }
}
class pair
{
 
  pair1 pr;
  int second;
 
  pair(pair1 pr, int second)
  {
    this.pr = pr;
    this.second = second;
  }
}
class pair1
{
  int a;
  int b;
  pair1(int a, int b)
  {
    this.a = a;
    this.b = b;
  }
}
 
// This code is contributed by aditya7409

Python3
# Python3 implementation to multiply
# the given subarray by the x
# for multiple queries of the Q
  
# Function to answer each query
def query(pre_sum, n, l, r, val):
 
    ans = 0
    temp = 0;
    if (l > 0):
        temp = pre_sum[l - 1];
     
    ans = val * (pre_sum[r] - temp);
    return ans;
   
# Function to multiply the subarray
# by the x for multiple Queries
def multiplyArray(arr, q, n):
     
    pre_sum = []
    s = 0;
      
    # Loop to compute the prefix
    # sum of the array
    for i in range(n):
 
        s += arr[i];
        pre_sum.append(s);
          
    # Loop to answer each query
    for i in q:
        print(query(pre_sum, n, i[0][0], i[0][1], i[1]))
         
# Driver Code
if __name__=='__main__':
 
    # Array
    arr = [ 2, 3, 4, 2, 5, 1 ]
    n = 6;
    q = []
    q.append([[2, 4], 5])
    q.append([[1, 1], 4])
    q.append([[1, 3], -2])
      
    # Function Call
    multiplyArray(arr, q, n);
      
# this code is contribute by rutvik_56

C#
// C# implementation to multiply
// the given subarray by the x
// for multiple queries of the Q
using System;
using System.Collections.Generic;
class GFG
{
     
    // Function to answer each query
    static int query(List pre_sum, int n,
             int l, int r, int val)
    {
        int ans;
        int temp = 0;
        if (l > 0)
        {
            temp = pre_sum[l - 1];
        }
        ans = val * (pre_sum[r] - temp);
        return ans;
    }
      
    // Function to multiply the subarray
    // by the x for multiple Queries
    static void multiplyArray(int[] arr, List, int>> q, int n)
    {
        List pre_sum = new List();
        int s = 0;
          
        // Loop to compute the prefix
        // sum of the array
        for (int i = 0; i < n; i++)
        {
            s += arr[i];
            pre_sum.Add(s);
        }
          
        // Loop to answer each query
        foreach(Tuple, int> i in q)
        {
            Console.WriteLine(query(pre_sum, n, i.Item1.Item1,
                                    i.Item1.Item2, i.Item2));
        }
    }
 
  // Driver code
  static void Main()
  {
     
    // Array
    int[] arr = { 2, 3, 4, 2, 5, 1 };
    int n = 6;
    List, int>> q = new List, int>>();
    q.Add(new Tuple, int>(new Tuple(2,4), 5));
    q.Add(new Tuple, int>(new Tuple(1,1), 4));
    q.Add(new Tuple, int>(new Tuple(1,3), -2));
      
    // Function Call
    multiplyArray(arr, q, n);
  }
}
 
// This code is contributed by divyeshrabadiya07

输出： 
55
12
-18