通过添加或相乘数组元素来更新数组并打印指定索引处的元素的查询

给定一个由N 个整数组成的数组arr[]和一个由M对表示{X, Y}类型查询的数组Q[] ，任务是执行以下类型的查询：

• Query(0, X)：将整数X添加到每个数组元素。
• Query(1, Y)：将每个数组元素乘以Y 。
• Query(2, i)：打印索引i处的值。

例子：

Input: arr[] = {5, 1, 2, 3, 9}, Q[][] = {{0, 5}, {2, 4}, {1, 4}, {0, 2}, {2, 3}}
Output: 14 34
Explanation:
Below are the results after performing each query:

1. Query(0, 5): Adding 5 to every array elements, modifies the array as arr[] = {10, 6, 7, 8, 14}.
2. Query(2, 4): Print the array element at index 4, i.e., arr[4] = 14.
3. Query(1, 4): Multiplying every array elements with 4, modifies the array as arr[] = {40, 24, 28, 32, 56}
4. Query(0, 2): Adding 2 to every array elements, modifies the array as arr[] = {42, 26, 30, 34, 58}
5. Query(2, 3): Print the element at index 4, i.e., arr[3] = 34.

Input: arr[] = {3, 1, 23, 45, 100}, Q[][] = {{1, 2}, {0, 10}, {2, 3}, {1, 5}, {2, 4}}
Output: 100 1050

朴素方法：解决给定问题的最简单方法是通过遍历给定数组来执行每个查询，并为每个查询打印相应的结果。

时间复杂度： O(N * M)
辅助空间： O(1)

高效的方法：上述方法可以基于以下观察进行优化：

• 假设A是一个元素，执行以下操作：
  • 加入A1至A，则A的值变为A = A + A1。
  • 将A乘以b1 ，则A的值变为A = b1 * A + b1 * a1 。
  • A2添加到A，则A的值变为A = B1 * A + B1 * A1 + A2。
  • 将A乘以b2 ，则A的值变为， A = b1 * b2 * A + b1 * b2 * a1 + a2 * b2
• 假设Mul是类型1查询的所有整数的乘积，而Add存储所有类型 0查询的值，并且类型 1以从0开始的给定顺序执行。那么从上面可以看出，任何数组元素arr[i] 都修改为(arr[i] * Mul + Add) 。

请按照以下步骤解决问题：

• 初始化两个变量，比如Mul为1和Add为0，存储类型2查询中所有整数的乘法，并存储在数字0上按给定顺序执行类型1和2查询后得到的值。
• 遍历给定的查询数组Q[][2]并执行以下步骤：
  • 如果Q[i][0]为0 ，则将Add的值增加Q[i][1] 。
  • 否则，如果Q[i][0] 的值为1，则将Mul和Add乘以Q[i][1] 。
  • 否则，打印值(arr[Q[i][1]] * Mul + Add)作为类型 2查询的结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to modify the array
// by performing given queries
void Query(int arr[], int N,
           vector > Q)
{
    // Stores the multiplication
    // of all integers of type 1
    int mul = 1;
 
    // Stores the value obtained after
    // performing queries of type 1 & 2
    int add = 0;
 
    // Iterate over the queries
    for (int i = 0; i < Q.size(); i++) {
 
        // Query of type 0
        if (Q[i][0] == 0) {
 
            // Update the value of add
            add = add + Q[i][1];
        }
 
        // Query of type 1
        else if (Q[i][0] == 1) {
 
            // Update the value of mul
            mul = mul * Q[i][1];
 
            // Update the value of add
            add = add * Q[i][1];
        }
 
        // Otherwise
        else {
 
            // Store the element at index Q[i][1]
            int ans = arr[Q[i][1]] * mul + add;
 
            // Print the result for
            // the current query
            cout << ans << " ";
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 1, 23, 45, 100 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    vector > Q = {
        { 1, 2 }, { 0, 10 },
        { 2, 3 }, { 1, 5 }, { 2, 4 }
    };
    Query(arr, N, Q);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG
{
 
    // Function to modify the array
    // by performing given queries
    static void Query(int arr[], int N, int Q[][])
    {
       
        // Stores the multiplication
        // of all integers of type 1
        int mul = 1;
 
        // Stores the value obtained after
        // performing queries of type 1 & 2
        int add = 0;
 
        // Iterate over the queries
        for (int i = 0; i < Q.length; i++) {
 
            // Query of type 0
            if (Q[i][0] == 0) {
 
                // Update the value of add
                add = add + Q[i][1];
            }
 
            // Query of type 1
            else if (Q[i][0] == 1) {
 
                // Update the value of mul
                mul = mul * Q[i][1];
 
                // Update the value of add
                add = add * Q[i][1];
            }
 
            // Otherwise
            else {
 
                // Store the element at index Q[i][1]
                int ans = arr[Q[i][1]] * mul + add;
 
                // Print the result for
                // the current query
                System.out.print(ans + " ");
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int arr[] = { 3, 1, 23, 45, 100 };
        int N = arr.length;
 
        int Q[][] = { { 1, 2 },
                      { 0, 10 },
                      { 2, 3 },
                      { 1, 5 },
                      { 2, 4 } };
        Query(arr, N, Q);
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to modify the array
# by performing given queries
def Query(arr, N, Q):
   
    # Stores the multiplication
    # of all integers of type 1
    mul = 1
 
    # Stores the value obtained after
    # performing queries of type 1 & 2
    add = 0
 
    # Iterate over the queries
    for i in range(len(Q)):
       
        # Query of type 0
        if (Q[i][0] == 0):
           
            # Update the value of add
            add = add + Q[i][1]
             
        # Query of type 1
        elif (Q[i][0] == 1):
 
            # Update the value of mul
            mul = mul * Q[i][1]
 
            # Update the value of add
            add = add * Q[i][1]
        # Otherwise
        else:
 
            # Store the element at index Q[i][1]
            ans = arr[Q[i][1]] * mul + add
 
            # Prthe result for
            # the current query
            print(ans,end=" ")
 
# Driver Code
if __name__ == '__main__':
 
    arr = [3, 1, 23, 45, 100]
    N = len(arr)
    Q = [
        [ 1, 2 ], [ 0, 10 ],
        [ 2, 3 ], [ 1, 5 ], [ 2, 4 ]
    ]
    Query(arr, N, Q)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to modify the array
  // by performing given queries
  static void Query(int[] arr, int N, int[, ] Q)
  {
    // Stores the multiplication
    // of all integers of type 1
    int mul = 1;
 
    // Stores the value obtained after
    // performing queries of type 1 & 2
    int add = 0;
 
    // Iterate over the queries
    for (int i = 0; i < Q.GetLength(0); i++) {
 
      // Query of type 0
      if (Q[i, 0] == 0) {
 
        // Update the value of add
        add = add + Q[i, 1];
      }
 
      // Query of type 1
      else if (Q[i, 0] == 1) {
 
        // Update the value of mul
        mul = mul * Q[i, 1];
 
        // Update the value of add
        add = add * Q[i, 1];
      }
 
      // Otherwise
      else {
 
        // Store the element at index Q[i][1]
        int ans = arr[Q[i, 1]] * mul + add;
 
        // Print the result for
        // the current query
        Console.Write(ans + " ");
      }
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = { 3, 1, 23, 45, 100 };
    int N = arr.Length;
 
    int[, ] Q = { { 1, 2 },
                 { 0, 10 },
                 { 2, 3 },
                 { 1, 5 },
                 { 2, 4 } };
    Query(arr, N, Q);
  }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript implementation of the above approach
 
    // Function to modify the array
    // by performing given queries
    function Query(arr, N, Q)
    {
       
        // Stores the multiplication
        // of all letegers of type 1
        let mul = 1;
 
        // Stores the value obtained after
        // performing queries of type 1 & 2
        let add = 0;
 
        // Iterate over the queries
        for (let i = 0; i < Q.length; i++) {
 
            // Query of type 0
            if (Q[i][0] == 0) {
 
                // Update the value of add
                add = add + Q[i][1];
            }
 
            // Query of type 1
            else if (Q[i][0] == 1) {
 
                // Update the value of mul
                mul = mul * Q[i][1];
 
                // Update the value of add
                add = add * Q[i][1];
            }
 
            // Otherwise
            else {
 
                // Store the element at index Q[i][1]
                let ans = arr[Q[i][1]] * mul + add;
 
                // Prlet the result for
                // the current query
                document.write(ans + " ");
            }
        }
    }
 
  // Driver Code
     
     let arr = [ 3, 1, 23, 45, 100 ];
        let N = arr.length;
 
        let Q = [[ 1, 2 ],
                      [ 0, 10 ],
                      [ 2, 3 ],
                      [ 1, 5 ],
                      [ 2, 4 ]];
        Query(arr, N, Q);
       

输出：

100 1050

时间复杂度： O(M)
辅助空间： O(1)

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