打印给定字符中主要索引处的字符串的程序

给定一个字符串，我们的任务是打印出现在素数索引处的字符。

例子 ：

Input : I love programming 
Output : lv gan
Explanation :
prime index characters in a string are : lv gan
          

Input : Happy coding everyone
Output : apycn ro

方法 ：

1. 使用两个循环将数字除以字符串的长度值。
2. 当余数为 0 时递增变量结果。
3. 如果变量 result = 1，则打印相应的字符。

以下是上述方法的实现：

C++

// C++ Program to print Characters at
// Prime index in a given String
#include 
using namespace std;
 
bool isPrime(int n)
{
    // Corner case
    if (n <= 1) return false;
 
    // Check from 2 to n-1
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function to print
// character at prime index
void prime_index(string input)
{
    int n = input.length();
 
    // Loop to check if
    // index prime or not
    for (int i = 2; i <= n; i++)
        if (isPrime(i))
            cout << input[i - 1];        
}
 
// Driver Code
int main()
{
    string input = "GeeksforGeeks";
    prime_index(input);
    return 0;
}

Java

// Java Program to print
// Characters at Prime index
// in a given String
class GFG
{
    static boolean isPrime(int n)
    {
        // Corner case
        if (n <= 1) return false;
     
        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;
     
        return true;
    }
     
    // Function to print
    // character at prime index
    static void prime_index(String input)
    {
        int n = input.length();
     
        // Loop to check if
        // index prime or not
        for (int i = 2; i <= n; i++)
            if (isPrime(i))
 
                System.out.print
                (input.charAt(i - 1));        
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String input = "GeeksforGeeks";
         
        prime_index(input);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to print
# Characters at Prime index
# in a given String
 
def isPrime(n):
 
    # Corner case
    if n <= 1:
        return False
 
    # Check from 2 to n-1
    for i in range(2, n):
        if n % i == 0:
            return False;
 
    return True
 
# Function to print
# character at prime index
def prime_index (input):
    p = list(input)
    s = ""
     
    # Loop to check if
    # index prime or not
    for i in range (2, len(p) + 1):
        if isPrime(i):
            s = s + input[i-1]
    print (s)
         
# Driver Code
input = "GeeksforGeeks"
prime_index(input)

C#

// C# Program to print Characters
// at Prime index in a given String
using System;
 
class GFG
{
    static bool isPrime(int n)
    {
        // Corner case
        if (n <= 1) return false;
     
        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;
     
        return true;
    }
     
    // Function to print character
    // at prime index
    static void prime_index(string input)
    {
        int n = input.Length;
     
        // Loop to check if
        // index prime or not
        for (int i = 2; i <= n; i++)
            if (isPrime(i))
 
                 
                Console.Write(input[i - 1]);        
    }
     
    // Driver code
    public static void Main ()
    {
        string input = "GeeksforGeeks";
         
        prime_index(input);
    }
}
 
// This code is contributed by Vt_m.

Javascript

 
// Javascript program to print characters
// at Prime index in a given String
function isPrime(n)
{
     
    // Corner case
    if (n <= 1) return false;
   
    // Check from 2 to n-1
    for(let i = 2; i < n; i++)
        if (n % i == 0)
            return false;
   
    return true;
}
   
// Function to print character
// at prime index
function prime_index(input)
{
    let n = input.length;
   
    // Loop to check if
    // index prime or not
    for(let i = 2; i <= n; i++)
        if (isPrime(i))
            document.write(input[i - 1]);        
}
 
// Driver code
let input = "GeeksforGeeks";
       
prime_index(input);
 
// This code is contributed by suresh07 
 

输出 ：

eesoes

优化：
对于大字符串，我们可以使用埃拉托色尼筛法有效地找到所有小于或等于字符串长度的素数。