数组上的乘法：O(1)中的范围更新查询

考虑一个整数数组A []和以下两种类型的查询。

1. update(l，r，x)：将x乘以从A [l]到A [r]的所有值(包括两端值)。
2. printArray()：打印当前修改后的数组。

例子：

Input: A[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
        update(0, 2, 2)
        update(1, 4, 3)
        print()
        update(4, 8, 5)
        print()
Output: 2 6 6 3 15 5 5 5 5 1
Explanation: 
The query update(0, 2, 2) 
multiply 2 to A[0], A[1] and A[2]. 
After update, A[] becomes {2, 2, 2, 1, 1, 1, 1, 1, 1, 1}       
Query update(1, 4, 3) multiply 3 to A[1],
A[2], A[3] and A[4]. After update, A[] becomes
{2, 6, 6, 3, 3, 1, 1, 1, 1, 1}.
Query update(4, 8, 5) multiply 5, A[4] to A[8]. 
After update, A[] becomes {2, 6, 6, 3, 15, 5, 5, 5, 5, 1}.

Input: A[] = {10, 5, 20, 40}
        update(0, 1, 10)
        update(1, 3, 20)
        update(2, 2, 2)
        print()
Output: 100 1000 800 800

方法：

一个简单的解决方案是执行以下操作：

1. update(l，r，x)：运行从l到r的循环，并将x乘以从A [l]到A [r]的所有元素。
2. print()：只需打印A []。

以上两个操作的时间复杂度均为O(n)。

高效方法：
一种有效的解决方案是使用两个数组，一个用于乘法，另一个用于除法。 mul []和div []。

1. 将x乘以mul [l]，将x乘以div [r + 1]
2. 采取mul数组mul [i] =(mul [i] * mul [i-1])/ div [i]的前缀乘法
3. printArray()：执行A [0] = mul [0]并打印。对于其余元素，请执行A [i] =(A [i] * mul [i])

下面是上述方法的实现：

C++

// C++ program for
// the above approach
#include 
using namespace std;
  
// Creates a mul[] array for A[] and returns
// it after filling initial values.
void initialize(int mul[], int div[], int size)
{
  
    for (int i = 1; i < size; i++) {
        mul[i] = (mul[i] * mul[i - 1]) / div[i];
    }
}
  
// Does range update
void update(int l, int r, int x, int mul[], int div[])
{
    mul[l] *= x;
    div[r + 1] *= x;
}
  
// Prints updated Array
void printArray(int ar[], int mul[], int div[], int n)
{
  
    for (int i = 0; i < n; i++) {
        ar[i] = ar[i] * mul[i];
        cout << ar[i] << " ";
    }
}
  
// Driver code;
int main()
{
  
    // Array to be updated
    int ar[] = { 10, 5, 20, 40 };
    int n = sizeof(ar) / sizeof(ar[0]);
  
    // Create and fill mul and div Array
    int mul[n + 1], div[n + 1];
  
    for (int i = 0; i < n + 1; i++) {
        mul[i] = div[i] = 1;
    }
  
    update(0, 1, 10, mul, div);
    update(1, 3, 20, mul, div);
    update(2, 2, 2, mul, div);
  
    initialize(mul, div, n + 1);
  
    printArray(ar, mul, div, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Creates a mul[] array for A[] and returns
// it after filling initial values.
static void initialize(int mul[], 
                       int div[], int size)
{
  
    for (int i = 1; i < size; i++) 
    {
        mul[i] = (mul[i] * mul[i - 1]) / div[i];
    }
}
  
// Does range update
static void update(int l, int r, int x, 
                   int mul[], int div[])
{
    mul[l] *= x;
    div[r + 1] *= x;
}
  
// Prints updated Array
static void printArray(int ar[], int mul[], 
                       int div[], int n)
{
    for (int i = 0; i < n; i++) 
    {
        ar[i] = ar[i] * mul[i];
        System.out.print(ar[i] + " ");
    }
}
  
// Driver code;
public static void main(String[] args)
{
    // Array to be updated
    int ar[] = { 10, 5, 20, 40 };
    int n = ar.length;
  
    // Create and fill mul and div Array
    int []mul = new int[n + 1];
    int []div = new int[n + 1];
  
    for (int i = 0; i < n + 1; i++)
    {
        mul[i] = div[i] = 1;
    }
  
    update(0, 1, 10, mul, div);
    update(1, 3, 20, mul, div);
    update(2, 2, 2, mul, div);
  
    initialize(mul, div, n + 1);
  
    printArray(ar, mul, div, n);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
  
# Creates a mul[] array for A[] and returns
# it after filling initial values.
def initialize(mul, div, size):
  
    for i in range(1, size):
        mul[i] = (mul[i] * mul[i - 1]) / div[i];
  
# Does range update
def update(l, r, x, mul, div):
    mul[l] *= x;
    div[r + 1] *= x;
  
# Prints updated Array
def printArray(ar, mul, div, n):
  
    for i in range(n):
        ar[i] = ar[i] * mul[i];
        print(int(ar[i]), end = " ");
  
# Driver code;
if __name__ == '__main__':
      
    # Array to be updated
    ar = [ 10, 5, 20, 40 ];
    n = len(ar);
  
    # Create and fill mul and div Array
    mul = [0] * (n + 1);
    div = [0] * (n + 1);
  
    for i in range(n + 1):
        mul[i] = div[i] = 1;
  
    update(0, 1, 10, mul, div);
    update(1, 3, 20, mul, div);
    update(2, 2, 2, mul, div);
  
    initialize(mul, div, n + 1);
  
    printArray(ar, mul, div, n);
  
# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;
  
class GFG 
{
  
// Creates a mul[] array for A[] and returns
// it after filling initial values.
static void initialize(int []mul, 
                       int []div, int size)
{
  
    for (int i = 1; i < size; i++) 
    {
        mul[i] = (mul[i] * mul[i - 1]) / div[i];
    }
}
  
// Does range update
static void update(int l, int r, int x, 
                   int []mul, int []div)
{
    mul[l] *= x;
    div[r + 1] *= x;
}
  
// Prints updated Array
static void printArray(int []ar, int []mul, 
                       int []div, int n)
{
    for (int i = 0; i < n; i++) 
    {
        ar[i] = ar[i] * mul[i];
        Console.Write(ar[i] + " ");
    }
}
  
// Driver code;
public static void Main(String[] args)
{
    // Array to be updated
    int []ar = { 10, 5, 20, 40 };
    int n = ar.Length;
  
    // Create and fill mul and div Array
    int []mul = new int[n + 1];
    int []div = new int[n + 1];
  
    for (int i = 0; i < n + 1; i++)
    {
        mul[i] = div[i] = 1;
    }
  
    update(0, 1, 10, mul, div);
    update(1, 3, 20, mul, div);
    update(2, 2, 2, mul, div);
  
    initialize(mul, div, n + 1);
  
    printArray(ar, mul, div, n);
}
}
  
// This code is contributed by Rajput-Ji

输出：

100 1000 800 800