GCD = 1的子数组数|段树

给定一个数组arr [] ，任务是查找GCD等于1的子数组的数量。

例子：

Input: arr[] = {1, 1, 1}
Output: 6
Every single subarray of the given array has GCD
of 1 and there are a total of 6 subarrays.

Input: arr[] = {2, 2, 2}
Output: 0

方法：可以使用段树数据结构在O(NlogN)中解决此问题。将要构建的段可用于回答range-gcd查询。

现在让我们了解一下算法。使用两指针技术可以解决此问题。在讨论该算法之前，让我们进行一些观察。

• 假设G是子数组arr [l … r]的GCD，而G1是子数组arr [l + 1 … r]的GCD。 G始终小于或等于G1 。
• 比方说，对于给定的L1，R1是第一索引，使得在范围[L，R]的GCD是1，那么对于任何L2大于或等于L1，R2也将大于或等于R1。

经过以上观察，两指针技术很有意义，即如果长度
对于索引L已知最小R的R个索引，然后对于索引L +1已知搜索需要从R开始。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define maxLen 30
  
// Array to store segment-tree
int seg[3 * maxLen];
  
// Function to build segment-tree to
// answer range GCD queries
int build(int l, int r, int in, int* arr)
{
    // Base-case
    if (l == r)
        return seg[in] = arr[l];
  
    // Mid element of the range
    int mid = (l + r) / 2;
  
    // Merging the result of left and right sub-tree
    return seg[in] = __gcd(build(l, mid, 2 * in + 1, arr),
                           build(mid + 1, r, 2 * in + 2, arr));
}
  
// Function to perform range GCD queries
int query(int l, int r, int l1, int r1, int in)
{
    // Base-cases
    if (l1 <= l and r <= r1)
        return seg[in];
    if (l > r1 or r < l1)
        return 0;
  
    // Mid-element
    int mid = (l + r) / 2;
  
    // Calling left and right child
    return __gcd(query(l, mid, l1, r1, 2 * in + 1),
                 query(mid + 1, r, l1, r1, 2 * in + 2));
}
  
// Function to find the required count
int findCnt(int* arr, int n)
{
    // Building the segment tree
    build(0, n - 1, 0, arr);
  
    // Two pointer variables
    int i = 0, j = 0;
  
    // To store the final answer
    int ans = 0;
  
    // Looping
    while (i < n) {
  
        // Incrementing j till we don't get
        // a gcd value of 1
        while (j < n and query(0, n - 1, i, j, 0) != 1)
            j++;
  
        // Updating the final answer
        ans += (n - j);
  
        // Increment i
        i++;
  
        // Update j
        j = max(j, i);
    }
  
    // Returning the final answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 1 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << findCnt(arr, n);
  
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
static int maxLen = 30;
  
// Array to store segment-tree
static int []seg = new int[3 * maxLen];
  
// Function to build segment-tree to
// answer range GCD queries
static int build(int l, int r, 
                 int in, int[] arr)
{
    // Base-case
    if (l == r)
        return seg[in] = arr[l];
  
    // Mid element of the range
    int mid = (l + r) / 2;
  
    // Merging the result of left and right sub-tree
    return seg[in] = __gcd(build(l, mid, 2 * in + 1, arr),
                           build(mid + 1, r, 2 * in + 2, arr));
}
  
// Function to perform range GCD queries
static int query(int l, int r, int l1, 
                        int r1, int in)
{
    // Base-cases
    if (l1 <= l && r <= r1)
        return seg[in];
    if (l > r1 || r < l1)
        return 0;
  
    // Mid-element
    int mid = (l + r) / 2;
  
    // Calling left and right child
    return __gcd(query(l, mid, l1, r1, 2 * in + 1),
                 query(mid + 1, r, l1, r1, 2 * in + 2));
}
  
// Function to find the required count
static int findCnt(int[] arr, int n)
{
    // Building the segment tree
    build(0, n - 1, 0, arr);
  
    // Two pointer variables
    int i = 0, j = 0;
  
    // To store the final answer
    int ans = 0;
  
    // Looping
    while (i < n)
    {
  
        // Incrementing j till we don't get
        // a gcd value of 1
        while (j < n && query(0, n - 1, 
                                 i, j, 0) != 1)
            j++;
  
        // Updating the final answer
        ans += (n - j);
  
        // Increment i
        i++;
  
        // Update j
        j = Math.max(j, i);
    }
  
    // Returning the final answer
    return ans;
}
  
static int __gcd(int a, int b) 
{ 
    return b == 0 ? a : __gcd(b, a % b);     
}
  
// Driver code
public static void main(String []args) 
{
    int arr[] = { 1, 1, 1, 1 };
    int n = arr.length;
  
    System.out.println(findCnt(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the above approach 
from math import gcd
  
maxLen = 30;
  
# Array to store segment-tree 
seg = [0] * (3 * maxLen); 
  
# Function to build segment-tree to 
# answer range GCD queries 
def build(l, r, i, arr) :
  
    # Base-case 
    if (l == r) :
        seg[i] = arr[l]; 
        return seg[i];
  
    # Mid element of the range 
    mid = (l + r) // 2; 
  
    # Merging the result of left and right sub-tree 
    seg[i] = gcd(build(l, mid, 2 * i + 1, arr),
                 build(mid + 1, r, 2 * i + 2, arr)); 
    return seg[i];
  
# Function to perform range GCD queries 
def query(l, r, l1, r1, i) :
  
    # Base-cases 
    if (l1 <= l and r <= r1) :
        return seg[i]; 
          
    if (l > r1 or r < l1) :
        return 0; 
  
    # Mid-element 
    mid = (l + r) // 2; 
  
    # Calling left and right child 
    return gcd(query(l, mid, l1, r1, 2 * i + 1), 
               query(mid + 1, r, l1, r1, 2 * i + 2)); 
  
# Function to find the required count 
def findCnt(arr, n) : 
  
    # Building the segment tree 
    build(0, n - 1, 0, arr); 
  
    # Two pointer variables 
    i = 0; j = 0; 
  
    # To store the final answer 
    ans = 0; 
  
    # Looping 
    while (i < n) :
  
        # Incrementing j till we don't get 
        # a gcd value of 1 
        while (j < n and 
               query(0, n - 1, i, j, 0) != 1) :
            j += 1; 
  
        # Updating the final answer 
        ans += (n - j); 
  
        # Increment i 
        i += 1; 
  
        # Update j 
        j = max(j, i); 
  
    # Returning the final answer 
    return ans; 
  
# Driver code 
if __name__ == "__main__" :
  
    arr = [ 1, 1, 1, 1 ]; 
    n = len(arr); 
  
    print(findCnt(arr, n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
      
class GFG
{
static int maxLen = 30;
  
// Array to store segment-tree
static int []seg = new int[3 * maxLen];
  
// Function to build segment-tree to
// answer range GCD queries
static int build(int l, int r, 
                 int iN, int[] arr)
{
    // Base-case
    if (l == r)
        return seg[iN] = arr[l];
  
    // Mid element of the range
    int mid = (l + r) / 2;
  
    // Merging the result of left and right sub-tree
    return seg[iN] = __gcd(build(l, mid, 2 * iN + 1, arr),
                           build(mid + 1, r, 2 * iN + 2, arr));
}
  
// Function to perform range GCD queries
static int query(int l, int r, int l1, 
                        int r1, int iN)
{
    // Base-cases
    if (l1 <= l && r <= r1)
        return seg[iN];
    if (l > r1 || r < l1)
        return 0;
  
    // Mid-element
    int mid = (l + r) / 2;
  
    // Calling left and right child
    return __gcd(query(l, mid, l1, r1, 2 * iN + 1),
                 query(mid + 1, r, l1, r1, 2 * iN + 2));
}
  
// Function to find the required count
static int findCnt(int[] arr, int n)
{
    // Building the segment tree
    build(0, n - 1, 0, arr);
  
    // Two pointer variables
    int i = 0, j = 0;
  
    // To store the final answer
    int ans = 0;
  
    // Looping
    while (i < n)
    {
  
        // Incrementing j till we don't get
        // a gcd value of 1
        while (j < n && query(0, n - 1, 
                               i, j, 0) != 1)
            j++;
  
        // Updating the final answer
        ans += (n - j);
  
        // Increment i
        i++;
  
        // Update j
        j = Math.Max(j, i);
    }
  
    // Returning the final answer
    return ans;
}
  
static int __gcd(int a, int b) 
{ 
    return b == 0 ? a : __gcd(b, a % b);     
}
  
// Driver code
public static void Main(String []args) 
{
    int []arr = { 1, 1, 1, 1 };
    int n = arr.Length;
  
    Console.WriteLine(findCnt(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

10