小于N且数字总和大于N的数字总和的最大数字

给定一个整数N ，任务是找到小于N的最大数字，以使它的数字之和大于N的数字之和。如果不满足任何条件，则打印-1 。
例子：

Input: N = 100 
Output: 99 
99 is the largest number less than 100 sum of whose digits is greater than the sum of the digits of 100
Input: N = 49 
Output: -1 
 

方法：从N-1到1循环，然后检查任何数字的总和是否大于N的总和。满足条件的第一个数字是必需的数字。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the sum of the digits of n
int sumOfDigits(int n)
{
    int res = 0;
 
    // Loop for each digit of the number
    while (n > 0) {
        res += n % 10;
        n /= 10;
    }
 
    return res;
}
 
// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
int findNumber(int n)
{
 
    // Starting from n-1
    int i = n - 1;
 
    // Check until 1
    while (i > 0) {
 
        // If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n))
            return i;
        i--;
    }
 
    // If the condition is not satisfied
    return -1;
}
 
// Driver code
int main()
{
    int n = 824;
    cout << findNumber(n);
 
    return 0;
}

Java

//Java implementation of the approach
 
import java.io.*;
 
class GFG {
    // Function to return the sum of the digits of n
static int sumOfDigits(int n)
{
    int res = 0;
 
    // Loop for each digit of the number
    while (n > 0) {
        res += n % 10;
        n /= 10;
    }
 
    return res;
}
 
// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
static int findNumber(int n)
{
 
    // Starting from n-1
    int i = n - 1;
 
    // Check until 1
    while (i > 0) {
 
        // If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n))
            return i;
        i--;
    }
 
    // If the condition is not satisfied
    return -1;
}
 
// Driver code
    public static void main (String[] args) {
 
    int n = 824;
    System.out.println (findNumber(n));
    }
//This code is contributed by akt_mit   
}

Python3

# Python3 implementation of the approach
 
# Function to return the sum
# of the digits of n
def sumOfDigits(n) :
 
    res = 0;
 
    # Loop for each digit of the number
    while (n > 0) :
        res += n % 10
        n /= 10
 
    return res;
 
# Function to return the greatest
# number less than n such that
# the sum of its digits is greater
# than the sum of the digits of n
def findNumber(n) :
 
    # Starting from n-1
    i = n - 1;
 
    # Check until 1
    while (i > 0) :
 
        # If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n)) :
            return i
             
        i -= 1
 
    # If the condition is not satisfied
    return -1;
 
# Driver code
if __name__ == "__main__" :
     
    n = 824;
    print(findNumber(n))
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG
{
// Function to return the sum
// of the digits of n
static int sumOfDigits(int n)
{
    int res = 0;
 
    // Loop for each digit of
    // the number
    while (n > 0)
    {
        res += n % 10;
        n /= 10;
    }
 
    return res;
}
 
// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
static int findNumber(int n)
{
 
    // Starting from n-1
    int i = n - 1;
 
    // Check until 1
    while (i > 0)
    {
 
        // If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n))
            return i;
        i--;
    }
 
    // If the condition is
    // not satisfied
    return -1;
}
 
// Driver code
static public void Main ()
{
    int n = 824;
    Console.WriteLine (findNumber(n));
}
}
 
// This code is contributed by @Tushil

PHP

 0)
    {
        $res += $n % 10;
        $n /= 10;
    }
 
    return $res;
}
 
// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
function findNumber($n)
{
 
    // Starting from n-1
    $i = $n - 1;
 
    // Check until 1
    while ($i > 0)
    {
 
        // If i satisfies the given condition
        if (sumOfDigits($i) > sumOfDigits($n))
            return $i;
        $i--;
    }
 
    // If the condition is not satisfied
    return -1;
}
 
// Driver code
$n = 824;
 
echo findNumber($n);
     
// This code is contributed by Mukul singh
?>

Javascript

// javascript implementation of the approach
 
    // Function to return the sum of the digits of n
    function sumOfDigits(n)
    {
        var res = 0;
 
        // Loop for each digit of the number
        while (n > 0)
        {
            res += n % 10;
            n = parseInt(n/10);
        }
 
        return res;
    }
 
    // Function to return the greatest
    // number less than n such that
    // the sum of its digits is greater
    // than the sum of the digits of n
    function findNumber(n)
    {
 
        // Starting from n-1
        var i = n - 1;
 
        // Check until 1
        while (i > 0)
        {
 
            // If i satisfies the given condition
            if (sumOfDigits(i) > sumOfDigits(n))
                return i;
            i--;
        }
 
        // If the condition is not satisfied
        return -1;
    }
 
    // Driver code
    var n = 824;
    document.write(findNumber(n));
 
// This code is contributed by Princi Singh

输出：

819