由总和等于 N 的不同数字组成的最大数

给定一个正整数N ，任务是找到由不同数字组成的最大正数，其数字之和等于N 。如果不存在这样的数字，则打印“-1” 。

例子：

Input: N = 25
Output: 98710
Explanation:
The number 98710 is the largest number that contains only unique digits and the sum of its digits (9 + 8 + 7 + 1 + 0) is N(= 25).

Input: N = 50
Output: -1

方法：根据以下观察可以解决给定的问题：

• 如果 N 的值至少为 45：所需的结果是-1 ，因为可以使用非重复数字组成的最大数字是9876543210 ，其数字总和为45 。
• 对于 N 的所有其他值：要获得最大的数字，从最大的数字开始，即9 ，并不断递减它并将其添加到所需的数字。所需的数字必须在其末尾包含0 ，因为它会增加值而不更改数字总和。

请按照以下步骤解决问题：

• 如果给定的数字N大于45 ，则打印“-1” 。
• 否则，请执行以下步骤：
  • 初始化一个变量，比如num为0来存储所需的结果，以及一个变量，比如digit ，为9 。
  • 迭代一个循环，直到N和digit的值为正 并执行以下步骤：
    • 如果一个数字的值是至多N，由10然后乘以NUM，然后数字的值添加到Num和由数字递减的N的值。
    • 此外，将数字的值减少1 。
  • 将变量num乘以10并在末尾附加数字0 。
• 完成上述步骤后，将num的值打印为结果数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the largest positive
// number made up of distinct digits
// having the sum of its digits as N
long largestNumber(int N)
{
    // If given number is greater
    // than 45, print -1
    if (N > 45)
        return -1;
 
    // Store the required number and
    // the digit to be considered
    int num = 0, digit = 9;
 
    // Loop until N > 0 and digit > 0
    while (N > 0 && digit > 0) {
 
        // If the current digit is
        // at most N then, add it
        // to number num
        if (digit <= N) {
 
            // Update the value of
            // num
            num *= 10;
            num += digit;
 
            // Decrement N by digit
            N -= digit;
        }
 
        // Consider the next lower
        // digit
        digit -= 1;
    }
 
    // Add 0 at the end and return
    // the number num
    return num * 10;
}
 
// Driver Code
int main()
{
    int N = 25;
    cout << largestNumber(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to find the largest positive
// number made up of distinct digits
// having the sum of its digits as N
static long largestNumber(int N)
{
     
    // If given number is greater
    // than 45, print -1
    if (N > 45)
        return -1;
 
    // Store the required number and
    // the digit to be considered
    int num = 0, digit = 9;
 
    // Loop until N > 0 and digit > 0
    while (N > 0 && digit > 0)
    {
         
        // If the current digit is
        // at most N then, add it
        // to number num
        if (digit <= N)
        {
             
            // Update the value of
            // num
            num *= 10;
            num += digit;
 
            // Decrement N by digit
            N -= digit;
        }
 
        // Consider the next lower
        // digit
        digit -= 1;
    }
 
    // Add 0 at the end and return
    // the number num
    return num * 10;
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 25;
     
    System.out.print(largestNumber(N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python program for the above approach
 
# Function to find the largest positive
# number made up of distinct digits
# having the sum of its digits as N
def largestNumber(N):
     
    # If given number is greater
    # than 45, print -1
    if (N > 45):
        return -1
     
    # Store the required number and
    # the digit to be considered
    num = 0
    digit = 9
     
    # Loop until N > 0 and digit > 0
    while (N > 0 and digit > 0):
         
        # If the current digit is
        # at most N then, add it
        # to number num
        if (digit <= N):
             
            # Update the value of
            # num
            num *= 10
            num += digit
             
            # Decrement N by digit
            N -= digit
         
        # Consider the next lower
        # digit
        digit -= 1
     
    # Add 0 at the end and return
    # the number num
    return num * 10
 
# Driver Code
N = 25
print(largestNumber(N))
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the largest positive
// number made up of distinct digits
// having the sum of its digits as N
static long largestNumber(int N)
{
     
    // If given number is greater
    // than 45, print -1
    if (N > 45)
        return -1;
 
    // Store the required number and
    // the digit to be considered
    int num = 0, digit = 9;
 
    // Loop until N > 0 and digit > 0
    while (N > 0 && digit > 0)
    {
         
        // If the current digit is
        // at most N then, add it
        // to number num
        if (digit <= N)
        {
             
            // Update the value of
            // num
            num *= 10;
            num += digit;
 
            // Decrement N by digit
            N -= digit;
        }
 
        // Consider the next lower
        // digit
        digit -= 1;
    }
 
    // Add 0 at the end and return
    // the number num
    return num * 10;
}
 
// Driver Code
public static void Main()
{
    int N = 25;
     
    Console.Write(largestNumber(N));
}
}
 
// This code is contributed by ukasp

Javascript

 
// JavaScript program for the above approach
 
// Function to find the largest positive
// number made up of distinct digits
// having the sum of its digits as N
function largestNumber(N)
{
     
    // If given number is greater
    // than 45, prlet -1
    if (N > 45)
        return -1;
 
    // Store the required number and
    // the digit to be considered
    let num = 0, digit = 9;
 
    // Loop until N > 0 and digit > 0
    while (N > 0 && digit > 0)
    {
         
        // If the current digit is
        // at most N then, add it
        // to number num
        if (digit <= N)
        {
             
            // Update the value of
            // num
            num *= 10;
            num += digit;
 
            // Decrement N by digit
            N -= digit;
        }
 
        // Consider the next lower
        // digit
        digit -= 1;
    }
 
    // Add 0 at the end and return
    // the number num
    return num * 10;
}
 
// Driver Code
     let N = 25;
     
    document.write(largestNumber(N));
 

输出：

98710

时间复杂度： O(1)
辅助空间： O(1)