给定一个包含N 个整数的数组 A ，任务是根据以下条件计算数组中形成循环的元素的数量。
从索引i开始遍历数组并跳转到下一个连接的索引。如果j = GCD(i, A[i]) % N ，则有向边从 A 的索引 i 退出到索引 j 。如果按照描述的顺序遍历数组，再次访问索引 i，则称索引 i 在数组中形成一个循环。
例子：
Input: A = { 1, 1, 6, 2 }
Output: 2
Explanation:
Possible traversals with the given condition are:
0 -> 1 -> 1
1 -> 1
2 -> 2
3 -> 2 -> 2
Clearly, only vertices 1 and 2 forms a cycle.
Input: A = {0, 0, 0, 6}
Output: 4
Explanation:
Possible traversals with the given condition are:
0 -> 0
1 -> 1
2 -> 2
3 -> 3
Clearly, all the vertices forms a cycle.
方法：
为了解决上面提到的问题，我们必须假设每个索引代表图的一个节点。
- 如果 j = GCD(i, A[i]) % n，每个节点都有一条从 A 的索引 i 到索引 j 的单一有向边。如果遍历从节点 i 开始。
- 节点 i 将被称为此遍历的父节点，并且此父节点将分配给遍历期间访问过的所有节点。
- 在遍历图时，如果我们发现一个已经访问过的节点，并且该访问节点的父节点与遍历的父节点相同，则检测到一个新的循环。
- 现在，这个循环中的每个节点都将被计数，因为它们中的每一个都在形成循环。要计算此循环中的节点数，请从此节点启动另一个 DFS，直到不再访问该节点为止。
- 对图中的每个节点 i 重复此过程。
在最坏的情况下，每个节点最多将被遍历 3 次。因此解决方案具有线性时间复杂度。
下面是上述方法的实现：
C++
// C++ program to number of elements
// which form a cycle in an array
#include
using namespace std;
#define mp make_pair
#define pb push_back
#define mod 1000000007
// Function to count number of
// elements forming a cycle
int solve(int A[], int n)
{
int i, cnt = 0, j;
// Array to store parent
// node of traversal.
int parent[n];
// Array to determine
// whether current node
// is already counted
// in the cycle.
int vis[n];
// Initialize the arrays.
memset(parent, -1, sizeof(parent));
memset(vis, 0, sizeof(vis));
for (i = 0; i < n; i++) {
j = i;
// Check if current node is already
// traversed or not. If node is not
// traversed yet then parent value
// will be -1.
if (parent[j] == -1) {
// Traverse the graph until an
// already visited node is not
// found.
while (parent[j] == -1) {
parent[j] = i;
j = __gcd(j, A[j]) % n;
}
// Check parent value to ensure
// a cycle is present.
if (parent[j] == i) {
// Count number of nodes in
// the cycle.
while (!vis[j]) {
vis[j] = 1;
cnt++;
j = __gcd(j, A[j]) % n;
}
}
}
}
return cnt;
}
int main()
{
int A[] = { 1, 1, 6, 2 };
int n = sizeof(A) / sizeof(A[0]);
cout << solve(A, n);
return 0;
} Java
// Java program to number of elements
// which form a cycle in an array
import java.util.*;
class GFG{
static final int mod = 1000000007;
// Function to count number of
// elements forming a cycle
static int solve(int A[], int n)
{
int i, cnt = 0, j;
// Array to store parent
// node of traversal.
int []parent = new int[n];
// Array to determine
// whether current node
// is already counted
// in the cycle.
int []vis = new int[n];
// Initialize the arrays.
Arrays.fill(parent, -1);
Arrays.fill(vis, 0);
for(i = 0; i < n; i++)
{
j = i;
// Check if current node is already
// traversed or not. If node is not
// traversed yet then parent value
// will be -1.
if (parent[j] == -1)
{
// Traverse the graph until an
// already visited node is not
// found.
while (parent[j] == -1)
{
parent[j] = i;
j = __gcd(j, A[j]) % n;
}
// Check parent value to ensure
// a cycle is present.
if (parent[j] == i)
{
// Count number of nodes in
// the cycle.
while (vis[j] == 0)
{
vis[j] = 1;
cnt++;
j = __gcd(j, A[j]) % n;
}
}
}
}
return cnt;
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver code
public static void main(String[] args)
{
int A[] = { 1, 1, 6, 2 };
int n = A.length;
System.out.print(solve(A, n));
}
}
// This code is contributed by gauravrajput1Python3
# Python3 program to number of elements
# which form a cycle in an array
import math
mod = 1000000007
# Function to count number of
# elements forming a cycle
def solve(A, n):
cnt = 0
# Array to store parent
# node of traversal.
parent = [-1] * n
# Array to determine
# whether current node
# is already counted
# in the cycle.
vis = [0] * n
for i in range(n):
j = i
# Check if current node is already
# traversed or not. If node is not
# traversed yet then parent value
# will be -1.
if (parent[j] == -1):
# Traverse the graph until an
# already visited node is not
# found.
while (parent[j] == -1):
parent[j] = i
j = math.gcd(j, A[j]) % n
# Check parent value to ensure
# a cycle is present.
if (parent[j] == i):
# Count number of nodes in
# the cycle.
while (vis[j] == 0):
vis[j] = 1
cnt += 1
j = math.gcd(j, A[j]) % n
return cnt
# Driver code
A = [ 1, 1, 6, 2 ]
n = len(A)
print(solve(A, n))
# This code is contributed by sanjoy_62C#
// C# program to number of elements
// which form a cycle in an array
using System;
class GFG{
// Function to count number of
// elements forming a cycle
static int solve(int[] A, int n)
{
int i, cnt = 0, j;
// Array to store parent
// node of traversal.
int[] parent = new int[n];
// Array to determine
// whether current node
// is already counted
// in the cycle.
int[] vis = new int[n];
// Initialize the arrays.
Array.Fill(parent, -1);
Array.Fill(vis, 0);
for(i = 0; i < n; i++)
{
j = i;
// Check if current node is already
// traversed or not. If node is not
// traversed yet then parent value
// will be -1.
if (parent[j] == -1)
{
// Traverse the graph until an
// already visited node is not
// found.
while (parent[j] == -1)
{
parent[j] = i;
j = __gcd(j, A[j]) % n;
}
// Check parent value to ensure
// a cycle is present.
if (parent[j] == i)
{
// Count number of nodes in
// the cycle.
while (vis[j] == 0)
{
vis[j] = 1;
cnt++;
j = __gcd(j, A[j]) % n;
}
}
}
}
return cnt;
}
static int __gcd(int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
// Driver code
static void Main()
{
int[] A = { 1, 1, 6, 2 };
int n = A.Length;
Console.WriteLine(solve(A, n));
}
}
// This code is contributed by divyeshrabadiya07输出：
2
时间复杂度： O(N)
空间复杂度： O(N)
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