小于 N 的最大数字，数字和大于 N 的数字和
给定一个整数 N ，任务是找到小于 N 的最大数字，使其位数之和大于 N 的位数之和。如果任何数字都不满足条件，则打印 -1 。 举例:
输入: N = 100 输出: 99 99 是小于 100 的最大数字，其位数之和大于 100 的位数之和 输入: N = 49 输出: -1
进场:从 N-1 到 1 开始一个循环，检查任意数字的位数之和是否大于 N 的位数之和。满足条件的第一个数字是所需的数字。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)
//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
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
//PHP implementation of the approach
// Function to return the sum of
// the digits of n
function sumOfDigits($n)
{
$res = 0;
// Loop for each digit of the number
while ($n > 0)
{
$res += $n % 10;
$n /= 10;
}
return $res;
}
// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
function findNumber($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
?>
java 描述语言
<script>
// 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
</script>
Output:
819
时间复杂度: O(N)
辅助空间: O(1)
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