从给定序列 Y 生成序列 X 使得 Yi = gcd(X1, X2 , ... , Xi)

给定一个大小为N的序列Y ，其中：

Y_i = gcd(X₁, X₂, X₃, . . ., X_i ) of some sequence X. 

任务是找到这样的序列X ，如果任何这样的 X 是可能的。

注意：如果 X 存在，则 X 可能有多个值。生成任何一个就足够了。

例子：

Input: N = 2, Y = [4, 2]
Output: [4, 2]
Explanation: Y₀ = gcd(X₀) = X₀ = 4. And gcd(4, 2) = 2 = Y₁.

Input: [1, 3]
Output: -1
Explanation: No such sequence can be formed.

方法：可以根据以下观察解决问题：

• Y的第 i 个元素 = gcd(X ₁ , X ₂ , X ₃ , . . ., X _i ) 和 Y 的第 (i+1) 个元素 = gcd(X ₁ , X ₂ , X ₃ , . . ., X _i, X _i+1 )。
• 所以 Y 的第 (i+1) 个元素可以写为 gcd(gcd(X ₁ , X ₂ , X ₃ , . . ., X _i ), X _i+1 ) ——-(1) 即 gcd( a, b, c) = gcd( gcd(a, b), c)。
• 所以 Y 的第 (i+1) 个元素是等式 1 中的gcd( Y 的第 i 个元素，X(i+1)) 。
• 因此，Y 的第 (i+1) 个元素必须是 Y的第 i 个元素的因子，因为两个元素的 gcd 是两个元素的除数。
• 由于 Y 的第 (i+1) 个元素除以 Y 的第 i 个元素， gcd( 第 i 个元素，第 (i+1) 个元素)总是等于第 (i+1) 个元素。

按照下图进行操作：

Illustration:

For example: N = 2, Y = {4, 2}

• X[0] = Y[0] = 4 because any number is GCD of itself.
• Y[1] = GCD(X[0], X[1]). Now GCD(X[0]) = Y[0]. So Y[1] = GCD(Y[0], X[1]). Therefore Y[1] should be a factor of Y[0].
• In the given example 2 is a factor of 4. So forming a sequence X is possible and X[1] can be same as Y[1]. 
  Assigning X[1] = 2 satisfies GCD (X[0, X[1]) = Y[1] = 2.
• So the final sequence X is {4, 2}.

所以根据上面的观察，按照下面提到的步骤来解决问题：

1. 从序列的开头开始遍历 Y。
2. 如果在给定的序列 Y 中有任何第 (i+1) 个元素不分割第 i 个元素，则无法生成序列 X。
3. 如果第(i+1)个元素整除第i个元素，则存在一个序列X，由Y的第(i+1)个元素的上述结论可以找到
   是 gcd( Y 的第 i 个元素，X _i+1 ) 并且X _i+1 = Y _i+1是X _i+1的可能值。

下面是该方法的实现：

C++

// C++ Program to implement above approach
#include 
using namespace std;
 
// Euclidean theorem to find gcd
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to check if it is possible
// to generate lost sequence X
void checkforX(int Y[], int N)
{
    int cur_gcd = Y[0];
    bool Xexist = true;
 
    // Loop to check existence of X
    for (int i = 1; i < N; i++) {
        cur_gcd = gcd(cur_gcd, Y[i]);
        if (cur_gcd != Y[i]) {
            Xexist = false;
            break;
        }
    }
 
    // Sequence X is found
    if (Xexist) {
 
        // Loop to generate X
        for (int i = 0; i < N; i++)
            cout << Y[i] << ' ';
    }
 
    // Sequence X can't be generated
    else {
        cout << -1 << endl;
    }
}
 
// Driver code
int main()
{
    int N = 2;
 
    // Sequence Y of size 2
    int Y[] = { 4, 2 };
    checkforX(Y, N);
    return 0;
}

C

// C program to implement above approach
#include 
 
// Euclidean theorem to calculate gcd
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to check if it is possible
// to generate lost sequence X
void checkforX(int Y[], int N)
{
    int cur_gcd = Y[0];
    int Xexist = 1;
 
    // Loop to check existence of X
    for (int i = 1; i < N; i++) {
        cur_gcd = gcd(cur_gcd, Y[i]);
        if (cur_gcd != Y[i]) {
            Xexist = 0;
            break;
        }
    }
 
    // Sequence X is found
    if (Xexist) {
 
        // Loop to generate X
        for (int i = 0; i < N; i++)
            printf("%d ", Y[i]);
    }
 
    // Sequence X can't be generated
    else {
        printf("-1");
    }
}
 
// Driver code
int main()
{
    int N = 2;
 
    // Sequence Y of size 2
    int Y[] = { 4, 2 };
    checkforX(Y, N);
    return 0;
}

Java

// Java Program to implement above approach
import java.util.*;
 
class GFG{
 
// Euclidean theorem to find gcd
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to check if it is possible
// to generate lost sequence X
static void checkforX(int Y[], int N)
{
    int cur_gcd = Y[0];
    boolean Xexist = true;
 
    // Loop to check existence of X
    for (int i = 1; i < N; i++) {
        cur_gcd = gcd(cur_gcd, Y[i]);
        if (cur_gcd != Y[i]) {
            Xexist = false;
            break;
        }
    }
 
    // Sequence X is found
    if (Xexist) {
 
        // Loop to generate X
        for (int i = 0; i < N; i++)
            System.out.print(Y[i] +" ");
    }
 
    // Sequence X can't be generated
    else {
        System.out.print(-1 +"\n");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 2;
 
    // Sequence Y of size 2
    int Y[] = { 4, 2 };
    checkforX(Y, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
 
# Euclidean theorem to find gcd
def gcd(a, b):
    if (b == 0):
        return a
    return gcd(b, a % b)
 
# Function to check if it is possible
# to generate lost sequence X
def checkforX(Y, N):
    cur_gcd = Y[0]
    Xexist = True
 
    # Loop to check existence of X
    for i in range(1, N):
        cur_gcd = gcd(cur_gcd, Y[i])
        if (cur_gcd != Y[i]):
            Xexist = False
            break
 
    # Sequence X is found
    if (Xexist):
 
        # Loop to generate X
        for i in range(N):
            print(Y[i], end=' ')
 
    # Sequence X can't be generated
    else:
        print(-1)
 
# Driver code
N = 2
 
# Sequence Y of size 2
Y = [4, 2]
checkforX(Y, N)
 
# This code is contributed by gfgking

C#

// C# Program to implement above approach
using System;
class GFG
{
   
// Euclidean theorem to find gcd
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to check if it is possible
// to generate lost sequence X
static void checkforX(int []Y, int N)
{
    int cur_gcd = Y[0];
    bool Xexist = true;
 
    // Loop to check existence of X
    for (int i = 1; i < N; i++) {
        cur_gcd = gcd(cur_gcd, Y[i]);
        if (cur_gcd != Y[i]) {
            Xexist = false;
            break;
        }
    }
 
    // Sequence X is found
    if (Xexist) {
 
        // Loop to generate X
        for (int i = 0; i < N; i++)
            Console.Write(Y[i] + " ");
    }
 
    // Sequence X can't be generated
    else {
        Console.Write(-1);
    }
}
 
// Driver code
public static void Main()
{
    int N = 2;
 
    // Sequence Y of size 2
    int []Y = { 4, 2 };
    checkforX(Y, N);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code for the above approach
 
        // Euclidean theorem to find gcd
        function gcd(a, b) {
            if (b == 0)
                return a;
            return gcd(b, a % b);
        }
 
        // Function to check if it is possible
        // to generate lost sequence X
        function checkforX(Y, N) {
            let cur_gcd = Y[0];
            let Xexist = true;
 
            // Loop to check existence of X
            for (let i = 1; i < N; i++) {
                cur_gcd = gcd(cur_gcd, Y[i]);
                if (cur_gcd != Y[i]) {
                    Xexist = false;
                    break;
                }
            }
 
            // Sequence X is found
            if (Xexist) {
 
                // Loop to generate X
                for (let i = 0; i < N; i++)
                    document.write(Y[i] + ' ');
            }
 
            // Sequence X can't be generated
            else {
                document.write(-1 + '<br>');
            }
        }
 
        // Driver code
        let N = 2;
 
        // Sequence Y of size 2
        let Y = [4, 2];
        checkforX(Y, N);
 
  // This code is contributed by Potta Lokesh
    

输出

4 2 

时间复杂度： O(N)
辅助空间： O(1)