查找 Q 查询添加 AP 项后得到的 Array

给定一个由N个整数和Q个查询组成的数组A[] ，假设Query[][]的形式为{L, R, a, d} ，这样每个查询都表示无限 AP，其中第一项为a ，公差为d .任务是在执行给定查询后打印更新后的数组，以便对于每个查询{L, R, a, d}将等差数列的第 ( ⁱ – L + 1)项的值添加到A[i]中[L, R]范围内的每个索引i 。

例子：

Input: A[]= {5, 4, 2, 8}, Q = 2, Query[][] = {{1, 2, 1, 3}, {1, 4, 4, 1}}
Output: 10 13 8 15
Explanation:
Following are the queries performed:
Query 1: The arithmetic progression is {1, 4, 7, …}. After adding the first term of the progression to index 1 and the second term of the progression to index 2, the array modifies to {6, 8, 2, 8}.
Query 2: The arithmetic progression is {4, 5, 6, 7, 8, …}. After adding 4 to A[1], 5 to A[2], 6 to A[3] and 7 to A[4] the array modifies to {10, 13, 8 15}.
Therefore, the resultant array is {10, 13, 8 15}.

Input: A[] = {1, 2, 3, 4, 5}, Q = 3, Query[][] = {{1, 2, 1, 3}, {1, 3, 4, 1}, {1, 4, 1, 2}}
Output: 7 14 14 11 5

编程需要懂一点英语

方法：给定问题可以通过在每个操作中迭代范围[L, R]并在每个索引i处添加给定算术级数的相应项来解决。请按照以下步骤解决问题：

使用变量i遍历数组Query[][]并对每个查询{L, R, a, d}执行以下步骤：
- 使用变量j遍历数组A[]，在[L – 1, R – 1]范围内
  - 将a的值与A[j]的值相加。
  - 将a的值增加d 。
完成上述步骤后，将数组A[]打印为结果更新后的数组。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find array after performing
// the given query to the array elements
void addAP(int A[], int Q, int operations[2][4])
{
 
    // Traverse the given query
    for (int j = 0; j < 2; ++j)
    {
        int L = operations[j][0], R = operations[j][1], a = operations[j][2], d = operations[j][3];
        int curr = a;
 
        // Traverse the given array
        for(int i = L - 1; i < R; i++){
             
            // Update the value of A[i]
            A[i] += curr;
 
            // Update the value of curr
            curr += d;
        }
    }
     
    // Print the array elements
    for (int i = 0; i < 4; ++i)
        cout << A[i] << " ";
}
 
// Driver Code
int main() {
    int A[] = {5, 4, 2, 8};
    int Q = 2;
    int Query[2][4] = {{1, 2, 1, 3}, {1, 4, 4, 1}};
     
    // Function Call
    addAP(A, Q, Query);
    return 0;
}
 
// This code is contributed by shubhamsingh10.

C

// C program for the above approach
#include 
 
// Function to find array after performing
// the given query to the array elements
void addAP(int A[], int Q, int operations[2][4])
{
 
    // Traverse the given query
    for (int j = 0; j < 2; ++j)
    {
        int L = operations[j][0], R = operations[j][1], a = operations[j][2], d = operations[j][3];
        int curr = a;
 
        // Traverse the given array
        for(int i = L - 1; i < R; i++){
             
            // Update the value of A[i]
            A[i] += curr;
 
            // Update the value of curr
            curr += d;
        }
    }
     
    // Print the array elements
    for (int i = 0; i < 4; ++i)
        printf("%d ", A[i]);
}
 
// Driver Code
int main() {
    int A[] = {5, 4, 2, 8};
    int Q = 2;
    int Query[2][4] = {{1, 2, 1, 3}, {1, 4, 4, 1}};
     
    // Function Call
    addAP(A, Q, Query);
    return 0;
}
 
// This code is contributed by shubhamsingh10.

Java

// Java program for the above approach
class GFG {
 
    // Function to find array after performing
    // the given query to the array elements
    public static void addAP(int A[], int Q, int[][] operations)
    {
 
        // Traverse the given query
        for (int j = 0; j < 2; ++j) {
            int L = operations[j][0], R = operations[j][1],
          a = operations[j][2], d = operations[j][3];
            int curr = a;
 
            // Traverse the given array
            for (int i = L - 1; i < R; i++) {
 
                // Update the value of A[i]
                A[i] += curr;
 
                // Update the value of curr
                curr += d;
            }
        }
 
        // Print the array elements
        for (int i = 0; i < 4; ++i)
            System.out.print(A[i] + " ");
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int A[] = { 5, 4, 2, 8 };
        int Q = 2;
        int query[][] = { { 1, 2, 1, 3 }, { 1, 4, 4, 1 } };
 
        // Function Call
        addAP(A, Q, query);
    }
}
 
// This code is contributed by _saurabh_jaiswal

Python3

# Python program for the above approach
 
# Function to find array after performing
# the given query to the array elements
def addAP(A, Q, operations):
 
    # Traverse the given query
    for L, R, a, d in operations:
 
        curr = a
 
        # Traverse the given array
        for i in range(L-1, R):
           
            # Update the value of A[i]
            A[i] += curr
 
            # Update the value of curr
            curr += d
     
    # Print the array elements
    for i in A:
        print(i, end =' ')
 
 
# Driver Code
 
A = [5, 4, 2, 8]
Q = 2
Query = [(1, 2, 1, 3), (1, 4, 4, 1)]
 
# Function Call
addAP(A, Q, Query)

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find array after performing
// the given query to the array elements
public static void addAP(int[] A, int Q,
                         int[,] operations)
{
     
    // Traverse the given query
    for(int j = 0; j < 2; ++j)
    {
        int L = operations[j, 0], R = operations[j, 1],
            a = operations[j, 2], d = operations[j, 3];
        int curr = a;
 
        // Traverse the given array
        for(int i = L - 1; i < R; i++)
        {
             
            // Update the value of A[i]
            A[i] += curr;
 
            // Update the value of curr
            curr += d;
        }
    }
 
    // Print the array elements
    for(int i = 0; i < 4; ++i)
        Console.Write(A[i] + " ");
}
 
// Driver code
public static void Main(string[] args)
{
    int[] A = { 5, 4, 2, 8 };
    int Q = 2;
    int[,] query = { { 1, 2, 1, 3 },
                     { 1, 4, 4, 1 } };
 
    // Function Call
    addAP(A, Q, query);
}
}
 
// This code is contributed by avijitmondal1998

Javascript

输出

10 13 8 15

时间复杂度： O(N*Q)
辅助空间： O(1)