在将第一个或最后一个K元素乘以X后计算数组的GCD的查询

给定一个由N个正整数组成的数组arr []和一个{a，K，X}类型的2D数组query [] [] ，使得如果a的值为1 ，则将前K个数组元素乘以X。否则，将最后K个数组元素乘以X。任务是在对原始阵列执行每次查询之后，计算阵列的GCD。

例子：

Input: arr[] = {2, 3, 4, 8}, Queries[][3] = {{1, 2, 2}, {2, 4, 5}}
Output: 2 5
Explanation:
Query 1: The given query is {1, 2, 2}. After multiplying the first 2 array elements by 2, arr[] modifies to {4, 6, 4, 8}. GCD of the modified array is 2.
Query 2: The given query is {2, 4, 5}. After multiplying the last 4 elements array elements by 5, arr[] modifies to {10, 15, 20, 40}. GCD of the updated array is 5.

Input: arr[] = {4, 12, 4, 9}, Queries[][3] = {{1, 3, 3}, {2, 4, 1}}
Output: 3 1

天真的方法：最简单的方法是通过执行每个查询来更新给定的数组，然后找到更新后的数组的GCD。

时间复杂度： O(N * Q * log(M))，其中M是数组中存在的最大元素。
辅助空间： O(N)

高效的方法：可以通过存储给定数组的前缀和后缀GCD数组并按照以下步骤在O(1)时间内解决每个查询来优化上述方法：

• 初始化大小为N的数组prefix []和后缀[] ，以存储给定数组的前缀和后缀GCD数组。
• 从正面和背面遍历数组，并在每个索引处找到前缀和后缀GCD，并将其分别存储在prefix []和suffix []中。
• 现在，遍历数组query [] ，并对每个查询{a，K，X}执行以下操作：
  • 如果K的值为N，则打印前缀[N – 1] * X的结果。
  • 如果a的值为1，则找到前缀[K – 1] * X和后缀[K]的GCD作为结果，因为前缀[K – 1] * X是前K个数字和后缀[ ]的新GCD。 K + 1]是其余数组元素的GCD。
  • 如果a的值为2，则找到前缀[N – K – 1]的GCD和后缀[N – K] * X作为结果，因为前缀[N – K – 1] * X是的新GCD。前K个数字和后缀[N – K]是其余数组元素的GCD。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the GCD after performing
// each query on array elements
void findGCDQueries(int arr[], int N,
                    int Queries[][3],
                    int Q)
{
    // Stores prefix array and suffix
    // array
    int prefix[N], suffix[N];
 
    prefix[0] = arr[0];
    suffix[N - 1] = arr[N - 1];
 
    // Build prefix array
    for (int i = 1; i < N; i++) {
        prefix[i] = __gcd(prefix[i - 1],
                          arr[i]);
    }
 
    // Build suffix array
    for (int i = N - 2; i >= 0; i--) {
        suffix[i] = __gcd(suffix[i + 1],
                          arr[i]);
    }
 
    // Traverse queries array
    for (int i = 0; i < Q; i++) {
 
        int a = Queries[i][0];
        int K = Queries[i][1];
        int X = Queries[i][2];
 
        // Edge Case when update is
        // is required till the end
        if (K == N) {
            cout << prefix[N - 1] * X;
            continue;
        }
 
        // Edge Case when update is
        // is required till the front
        if (a == 1) {
            cout << __gcd(prefix[K - 1] * X,
                          suffix[K]);
        }
 
        // Find the resultant operation
        // for each query
        else {
            cout << __gcd(suffix[N - K] * X,
                          prefix[N - K - 1]);
        }
 
        cout << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 8 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int Queries[][3] = {
        { 1, 2, 2 },
        { 2, 4, 5 }
    };
    int Q = sizeof(Queries)
            / sizeof(Queries[0]);
 
    findGCDQueries(arr, N, Queries, Q);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG
{
 
  // Recursive function to return gcd of a and b
  static int gcd(int a, int b)
  {
 
    // Everything divides 0
    if (a == 0)
      return b;
    if (b == 0)
      return a;
 
    // base case
    if (a == b)
      return a;
 
    // a is greater
    if (a > b)
      return gcd(a - b, b);
    return gcd(a, b - a);
  }
 
 
  // Function to find the GCD after performing
  // each query on array elements
  static void findGCDQueries(int arr[], int N,
                             int Queries[][],
                             int Q)
  {
 
    // Stores prefix array and suffix
    // array
    int prefix[] = new int[N], suffix[] = new int[N];
 
    prefix[0] = arr[0];
    suffix[N - 1] = arr[N - 1];
 
    // Build prefix array
    for (int i = 1; i < N; i++) {
      prefix[i] = gcd(prefix[i - 1],
                      arr[i]);
    }
 
    // Build suffix array
    for (int i = N - 2; i >= 0; i--) {
      suffix[i] = gcd(suffix[i + 1],
                      arr[i]);
    }
 
    // Traverse queries array
    for (int i = 0; i < Q; i++) {
 
      int a = Queries[i][0];
      int K = Queries[i][1];
      int X = Queries[i][2];
 
      // Edge Case when update is
      // is required till the end
      if (K == N) {
        System.out.print(prefix[N - 1] * X);
        continue;
      }
 
      // Edge Case when update is
      // is required till the front
      if (a == 1) {
        System.out.print(gcd(prefix[K - 1] * X,
                             suffix[K]));
      }
 
      // Find the resultant operation
      // for each query
      else {
        System.out.print(gcd(suffix[N - K] * X,
                             prefix[N - K - 1]));
      }
 
      System.out.print(" ");
    }
  }
   
  // Driver Code
  public static void main(String[] args)
  {
 
    int arr[] = { 2, 3, 4, 8 };
    int N = arr.length;
    int Queries[][] = {
      { 1, 2, 2 },
      { 2, 4, 5 }
    };
    int Q = Queries.length;
 
    findGCDQueries(arr, N, Queries, Q);
  }
}
 
// This code is cntributed by sanjoy_62.

Python3

# Python 3 program for the above approach
from math import gcd
 
# Function to find the GCD after performing
# each query on array elements
def findGCDQueries(arr, N, Queries, Q):
   
    # Stores prefix array and suffix
    # array
    prefix = [0 for i in range(N)]
    suffix = [0 for i in range(N)]
 
    prefix[0] = arr[0]
    suffix[N - 1] = arr[N - 1]
 
    # Build prefix array
    for i in range(1,N,1):
        prefix[i] = gcd(prefix[i - 1], arr[i])
 
    # Build suffix array
    i = N - 2
    while(i>= 0):
        suffix[i] = gcd(suffix[i + 1], arr[i])
        i -= 1
 
    # Traverse queries array
    for i in range(Q):
        a = Queries[i][0]
        K = Queries[i][1]
        X = Queries[i][2]
 
        # Edge Case when update is
        # is required till the end
        if (K == N):
            print(prefix[N - 1] * X,end = " ")
            continue
 
        # Edge Case when update is
        # is required till the front
        if (a == 1):
            print(gcd(prefix[K - 1] * X,suffix[K]),end = " ")
 
        # Find the resultant operation
        # for each query
        else:
            print(gcd(suffix[N - K] * X, prefix[N - K - 1]),end = " ")
 
# Driver Code
if __name__ == '__main__':
    arr =  [2, 3, 4, 8]
    N = len(arr)
    Queries = [[1, 2, 2], [2, 4, 5]]
    Q = len(Queries)
    findGCDQueries(arr, N, Queries, Q)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program to implement
// the above approach
using System;
public class GFG
{
 
  // Recursive function to return gcd of a and b
  static int gcd(int a, int b)
  {
 
    // Everything divides 0
    if (a == 0)
      return b;
    if (b == 0)
      return a;
 
    // base case
    if (a == b)
      return a;
 
    // a is greater
    if (a > b)
      return gcd(a - b, b);
    return gcd(a, b - a);
  }
 
  // Function to find the GCD after performing
  // each query on array elements
  static void findGCDQueries(int []arr, int N,
                             int [,]Queries,
                             int Q)
  {
 
    // Stores prefix array and suffix
    // array
    int []prefix = new int[N];
    int []suffix = new int[N];
 
    prefix[0] = arr[0];
    suffix[N - 1] = arr[N - 1];
 
    // Build prefix array
    for (int i = 1; i < N; i++) {
      prefix[i] = gcd(prefix[i - 1],
                      arr[i]);
    }
 
    // Build suffix array
    for (int i = N - 2; i >= 0; i--) {
      suffix[i] = gcd(suffix[i + 1],
                      arr[i]);
    }
 
    // Traverse queries array
    for (int i = 0; i < Q; i++) {
 
      int a = Queries[i,0];
      int K = Queries[i,1];
      int X = Queries[i,2];
 
      // Edge Case when update is
      // is required till the end
      if (K == N) {
        Console.Write(prefix[N - 1] * X);
        continue;
      }
 
      // Edge Case when update is
      // is required till the front
      if (a == 1) {
        Console.Write(gcd(prefix[K - 1] * X,
                          suffix[K]));
      }
 
      // Find the resultant operation
      // for each query
      else {
        Console.Write(gcd(suffix[N - K] * X,
                          prefix[N - K - 1]));
      }
 
      Console.Write(" ");
    }
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    int []arr = { 2, 3, 4, 8 };
    int N = arr.Length;
    int [,]Queries = {
      { 1, 2, 2 },
      { 2, 4, 5 }
    };
    int Q = Queries.GetLength(0);
 
    findGCDQueries(arr, N, Queries, Q);
  }
}
 
// This code is cntributed by AnkThon.

输出

2 5

时间复杂度： O((N + Q)* log M)，其中M是数组的最大元素。
辅助空间： O(N)