通过翻转字符修改二进制字符串，以使由1组成的任何一对索引都不是互质的，也不能被彼此整除

给定一个整数N和一个最初由4 * N个0组成的二进制字符串，任务是翻转字符，使得由1 s组成的字符串的任何两个索引对都不是互素数，索引对也可以彼此可以被整除
注意：考虑基于1的索引。

例子：

Input: N = 3, S = “000000000000”
Output: 000000010101
Explanation: In the modified string “000000010101”, the indices of 1s are {8, 10, 12}. In the above set of indices, there does not exist any pair of indices that are co-prime and divisible by each other.

Input: N = 2, S = “00000000”
Output: 00000101

方法：给定的问题可以基于以下观察结果来解决：如果将字符翻转到位置4 * N，4 * N – 2，4 * N – 4，…最多N个词，则不存在任何项彼此可整除且GCD为1的一对索引。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
// Function to modify a string such
// that there doesn't exist any pair
// of indices consisting of 1s, whose
// GCD is 1 and are divisible by each other
void findString(char S[], int N)
{
  int strLen = 4 * N;
 
  // Flips characters at indices
  // 4N, 4N - 2, 4N - 4 .... upto N terms
  for (int i = 1; i <= N; i++) {
 
    S[strLen - 1] = '1';
    strLen -= 2;
  }
 
  // Print the string
  for (int i = 0; i < 4 * N; i++) {
    cout << S[i];
  }
}
 
// Driver code
int main()
{
 
  int N = 2;
 
  char S[4 * N];
 
  // Initialize the string S
  for (int i = 0; i < 4 * N; i++)
    S[i] = '0';
  // function call
  findString(S, N);
  return 0;
}
 
// This code is contributed by aditya7409.

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to modify a string such
    // that there doesn't exist any pair
    // of indices consisting of 1s, whose
    // GCD is 1 and are divisible by each other
    public static void findString(char S[], int N)
    {
        int strLen = 4 * N;
 
        // Flips characters at indices
        // 4N, 4N - 2, 4N - 4 .... upto N terms
        for (int i = 1; i <= N; i++) {
 
            S[strLen - 1] = '1';
            strLen -= 2;
        }
 
        // Print the string
        System.out.println(S);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 2;
 
        char S[] = new char[4 * N];
 
        // Initialize the string S
        for (int i = 0; i < 4 * N; i++)
            S[i] = '0';
 
        findString(S, N);
    }
}

Python3

# Python3 program for the above approach
 
# Function to modify a string such
# that there doesn't exist any pair
# of indices consisting of 1s, whose
# GCD is 1 and are divisible by each other
def findString(S, N) :
  strLen = 4 * N
 
  # Flips characters at indices
  # 4N, 4N - 2, 4N - 4 .... upto N terms
  for i in range(1, N + 1):
    S[strLen - 1] = '1'
    strLen -= 2
   
  # Prthe string
  for i in range(4 * N):
    print(S[i], end = "")
 
# Driver code
 
N = 2
S = [0] * (4 * N)
 
# Initialize the string S
for i in range(4 * N):
    S[i] = '0'
     
# function call
findString(S, N)
 
# This code is contributed by sanjoy_62.

C#

// C# program to implement
// the above approach
using System;
 
public class GFG
{
   
    // Function to modify a string such
    // that there doesn't exist any pair
    // of indices consisting of 1s, whose
    // GCD is 1 and are divisible by each other
    public static void findString(char[] S, int N)
    {
        int strLen = 4 * N;
 
        // Flips characters at indices
        // 4N, 4N - 2, 4N - 4 .... upto N terms
        for (int i = 1; i <= N; i++) {
 
            S[strLen - 1] = '1';
            strLen -= 2;
        }
 
        // Print the string
        Console.WriteLine(S);
    }
 
// Driver Code
public static void Main(String[] args)
{
    int N = 2;
    char[] S = new char[4 * N];
 
    // Initialize the string S
    for (int i = 0; i < 4 * N; i++)
        S[i] = '0';
 
    findString(S, N);
}
}
 
// This code is contributed by souravghosh0416.

输出：

00000101

时间复杂度： O(N)
辅助空间： O(1)