与字符的总和，按字典顺序最大的字符串等于n

给定一个正整数N ，任务是找到由小写英文字母组成的字典序最大的字符串，使得字符串的字符总和等于N其中‘a’ = 1 , ‘b’ = 2 , ‘c’ = 3 , ….. 和‘z’ = 26 。

例子：

Input: N = 30
Output: zd
Explanation:
The lexicographically largest string formed is “zd” whose sum of position of characters is (26 + 4) = 30(= N).

Input: N = 14
Output: n

方法：为了使字典顺序最大的字符串，这个想法是打印的字符Z，N / 26的次数，然后在(N％26 + 1)的英文字母位置ⁱ的字符。请按照以下步骤解决问题：

• 初始化一个字符串，比如ans ，它存储所需的字典序最大的字符串。
• 迭代直到N至少为26并执行以下步骤：
  • 将字符z添加到字符串ans 。
  • 将N的值减少26 。
• 将char(N + ‘a’)的值添加到字符串ans 。
• 完成上述步骤后，打印ans的值作为结果字符串。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to construct the
// lexicographically largest string
// having sum of characters as N
string getString(int N)
{
   
    // Stores the resulting string
    string ans = "";
 
    // Iterate until N is at least 26
    while (N >= 26) {
 
        // Append 'z' to the string ans
        ans += 'z';
 
        // Decrement N by 26
        N -= 26;
    }
 
    // Append character at index (N + 'a')
    ans += char(N + 'a' - 1);
 
    // Return the resultant string
    return ans;
}
 
// Driver Code
int main()
{
    int N = 30;
    cout << getString(N);
 
    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to construct the
    // lexicographically largest string
    // having sum of characters as N
    static String getString(int N)
    {
 
        // Stores the resulting string
        String ans = "";
 
        // Iterate until N is at least 26
        while (N >= 26) {
 
            // Append 'z' to the string ans
            ans += 'z';
 
            // Decrement N by 26
            N -= 26;
        }
 
        // Append character at index (N + 'a')
        ans += (char)(N + 'a' - 1);
 
        // Return the resultant string
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int N = 30;
        System.out.print(getString(N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to construct the
# lexicographically largest string
# having sum of characters as N
def getString(N):
   
    # Stores the resulting string
    ans = ""
 
    # Iterate until N is at least 26
    while (N >= 26):
 
        # Append 'z' to the string ans
        ans += 'z'
 
        # Decrement N by 26
        N -= 26
 
    # Append  character at index (N + 'a')
    ans += chr(N + ord('a') - 1)
 
    # Return the resultant string
    return ans
 
# Driver Code
if __name__ == '__main__':
    N = 30
    print(getString(N))
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
 
public class GFG {
 
    // Function to construct the
    // lexicographically largest string
    // having sum of characters as N
    static string getString(int N)
    {
 
        // Stores the resulting string
        string ans = "";
 
        // Iterate until N is at least 26
        while (N >= 26) {
 
            // Append 'z' to the string ans
            ans += 'z';
 
            // Decrement N by 26
            N -= 26;
        }
 
        // Append character at index (N + 'a')
        ans += (char)(N + 'a' - 1);
 
        // Return the resultant string
        return ans;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
 
        int N = 30;
        Console.WriteLine(getString(N));
    }
}
 
// This code is contributed by ukasp.

Javascript

    // Javascript program for the above approach
 
// Function to construct the
// lexicographically largest string
// having sum of characters as N
function getString(N){
 
    // Stores the resulting string
    let ans = ""
 
    // Iterate until N is at least 26
    while (N >= 26){
 
        // Append 'z' to the string ans
        ans += 'z'
 
        // Decrement N by 26
        N -= 26
    }
    // Append character at index (N + 'a')
    ans += String.fromCharCode(N + 'a'.charCodeAt(0) - 1)
 
    // Return the resultant string
    return ans
}
 
// Driver Code
    let N = 30
    document.write(getString(N))
 
// This code is contributed by Saurabh Jaiswal

输出：

zd

时间复杂度： O(N)
辅助空间： O(N)

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