最小的字符串可被两个给定的字符串整除

给定两个字符串s和长度为N T和M，任务是找到是双方两个字符串分割最小的字符串。如果不存在这样的字符串，则打印-1 。

For any two strings A and B, B divides A if and only if A is the concatenation of B at least once.

例子：

Input: S = “abab”, T = “ab”
Output: abab
Explanation: The string “abab” is same as S and twice the concatenation of string T (“abab” = “ab” + “ab” = T + T)

Input: S = “ccc”, T = “cc”
Output: cccccc
Explanation: The string “cccccc” is concatenation of S and T twice and thrice respectively.
(“cccccc” = “ccc” + “ccc” = S + S)
(“cccccc” = “cc” + “cc” + “cc” = T + T + T)

方法：该想法基于以下观察：所需字符串的长度(例如L)必须等于N和M的最小公倍数。检查字符串S的串联L / N次数是否等于字符串T的串联L / M次数。如果发现是真的，则打印其中任何一个。否则，打印-1 。请按照以下步骤解决问题：

• 将N和M的最小公倍数存储在变量中，例如L。
• 初始化两个字符串S1和T1 。
• 将字符串S1与字符串S (L / N)次数连接。
• 将字符串T1与字符串T (L / M)次数连接起来。
• 如果字符串S1和T1相等，则打印S1作为结果。
• 否则，打印-1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate
// GCD of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to calculate
// LCM of two numbers
int lcm(int a, int b)
{
    return (a / gcd(a, b)) * b;
}
 
// Function to find the smallest string
// which is divisible by strings S and T
void findSmallestString(string s, string t)
{
    // Store the length of both strings
    int n = s.length(), m = t.length();
 
    // Store LCM of n and m
    int l = lcm(n, m);
 
    // Temporary strings to store
    // concatenated strings
    string s1 = "", t1 = "";
 
    // Concatenate s1 (l / n) times
    for (int i = 0; i < l / n; i++) {
        s1 += s;
    }
 
    // Concatenate t1 (l / m) times
    for (int i = 0; i < l / m; i++) {
        t1 += t;
    }
 
    // If s1 and t1 are equal
    if (s1 == t1)
        cout << s1;
 
    // Otherwise, print -1
    else
        cout << -1;
}
 
// Driver Code
int main()
{
    string S = "baba", T = "ba";
    findSmallestString(S, T);
 
    return 0;
}

Java

// Java program for above approach
import java.io.*;
 
class GFG
{
 
  // Function to calculate
  // GCD of two numbers
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
 
  // Function to calculate
  // LCM of two numbers
  static int lcm(int a, int b)
  {
    return (a / gcd(a, b)) * b;
  }
 
  // Function to find the smallest string
  // which is divisible by strings S and T
  static void findSmallestString(String s, String t)
  {
    // Store the length of both strings
    int n = s.length(), m = t.length();
 
    // Store LCM of n and m
    int l = lcm(n, m);
 
    // Temporary strings to store
    // concatenated strings
    String s1 = "", t1 = "";
 
    // Concatenate s1 (l / n) times
    for (int i = 0; i < l / n; i++) {
      s1 += s;
    }
 
    // Concatenate t1 (l / m) times
    for (int i = 0; i < l / m; i++) {
      t1 += t;
    }
 
    // If s1 and t1 are equal
    if (s1.equals(t1)){
      System.out.println(s1);
    }
 
    // Otherwise, print -1
    else{
      System.out.println(-1);
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    String S = "baba", T = "ba";
    findSmallestString(S, T);
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Python3

# Python3 program for the above approach
 
# Function to calculate
# GCD of two numbers
def gcd(a, b):
    if (b == 0):
        return a
    return gcd(b, a % b)
 
# Function to calculate
# LCM of two numbers
def lcm(a, b):
    return (a // gcd(a, b)) * b
 
# Function to find the smallest string
# which is divisible by strings S and T
def findSmallestString(s, t):
    # Store the length of both strings
    n, m = len(s), len(t)
 
    # Store LCM of n and m
    l = lcm(n, m)
 
    # Temporary strings to store
    # concatenated strings
    s1, t1 = "", ""
 
    # Concatenate s1 (l / n) times
    for i in range(l//n):
        s1 += s
 
    # Concatenate t1 (l / m) times
    for i in range(l//m):
        t1 += t
 
    # If s1 and t1 are equal
    if (s1 == t1):
        print(s1)
 
    # Otherwise, pr-1
    else:
        print(-1)
 
# Driver Code
if __name__ == '__main__':
    S, T = "baba", "ba"
    findSmallestString(S, T)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for above approach
using System;
 
public class GFG
{
 
  // Function to calculate
  // GCD of two numbers
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
 
  // Function to calculate
  // LCM of two numbers
  static int lcm(int a, int b)
  {
    return (a / gcd(a, b)) * b;
  }
 
  // Function to find the smallest string
  // which is divisible by strings S and T
  static void findSmallestString(string s, string t)
  {
    // Store the length of both strings
    int n = s.Length, m = t.Length;
 
    // Store LCM of n and m
    int l = lcm(n, m);
 
    // Temporary strings to store
    // concatenated strings
    string s1 = "", t1 = "";
 
    // Concatenate s1 (l / n) times
    for (int i = 0; i < l / n; i++) {
      s1 += s;
    }
 
    // Concatenate t1 (l / m) times
    for (int i = 0; i < l / m; i++) {
      t1 += t;
    }
 
    // If s1 and t1 are equal
    if (s1 == t1)
      Console.WriteLine(s1);
 
    // Otherwise, print -1
    else
      Console.WriteLine(-1);
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    string S = "baba", T = "ba";
    findSmallestString(S, T);
  }
}
 
// This code is contributed by sanjoy_62.

输出：

baba

时间复杂度： O(max(N，M))
辅助空间： O(max(N，M))