# 算法：查找数字总和为N的最小数字

2021年3月25日12:33:12 发表评论 439 次浏览

## 本文概述

Input: N = 10
Output: 19
Explanation:
1 + 9 = 10 = N

Input: N = 18
Output: 99
Explanation:
9 + 9 = 18 = N

• 天真的方法是从0开始运行i的循环并查找数字总和的i并检查它是否等于N。

## C ++

// C++ program to find the smallest
// number whose sum of digits is also N
#include <iostream>
#include <math.h>
using namespace std;

// Function to get sum of digits
int getSum( int n)
{
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}

// Function to find the smallest
// number whose sum of digits is also N
void smallestNumber( int N)
{
int i = 1;
while (1) {
// Checking if number has
// sum of digits = N
if (getSum(i) == N) {
cout << i;
break ;
}
i++;
}
}

// Driver code
int main()
{
int N = 10;
smallestNumber(N);

return 0;
}

## Java

// Java program to find the smallest
// number whose sum of digits is also N
class GFG{

// Function to get sum of digits
static int getSum( int n)
{
int sum = 0 ;
while (n != 0 )
{
sum = sum + n % 10 ;
n = n / 10 ;
}
return sum;
}

// Function to find the smallest
// number whose sum of digits is also N
static void smallestNumber( int N)
{
int i = 1 ;
while ( 1 != 0 )
{
// Checking if number has
// sum of digits = N
if (getSum(i) == N)
{
System.out.print(i);
break ;
}
i++;
}
}

// Driver code
public static void main(String[] args)
{
int N = 10 ;
smallestNumber(N);
}
}

// This code is contributed
// by shivanisinghss2110

## Python3

# Python3 program to find the smallest
# number whose sum of digits is also N

# Function to get sum of digits
def getSum(n):

sum1 = 0 ;
while (n ! = 0 ):
sum1 = sum1 + n % 10 ;
n = n / / 10 ;

return sum1;

# Function to find the smallest
# number whose sum of digits is also N
def smallestNumber(N):

i = 1 ;
while ( 1 ):
# Checking if number has
# sum of digits = N
if (getSum(i) = = N):
print (i);
break ;

i + = 1 ;

# Driver code
N = 10 ;
smallestNumber(N);

# This code is contributed by Code_Mech

## C#

// C# program to find the smallest
// number whose sum of digits is also N
using System;

class GFG{

// Function to get sum of digits
static int getSum( int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n / 10;
}
return sum;
}

// Function to find the smallest
// number whose sum of digits is also N
static void smallestNumber( int N)
{
int i = 1;
while (1 != 0)
{

// Checking if number has
// sum of digits = N
if (getSum(i) == N)
{
Console.Write(i);
break ;
}
i++;
}
}

// Driver code
public static void Main(String[] args)
{
int N = 10;

smallestNumber(N);
}
}

// This code is contributed by Amit Katiyar

19

• 解决此问题的有效方法是观察。我们来看一些例子。
• 如果N = 10, 则ans = 19
• 如果N = 20, 则ans = 299
• 如果N = 30, 则ans = 3999
• 因此, 很明显答案将除第一个数字外的所有数字均为9, 因此我们得到的数字最小。
• 因此, 第N个项将是=

## C ++

// C++ program to find the smallest
// number whose sum of digits is also N
#include <iostream>
#include <math.h>
using namespace std;

// Function to find the smallest
// number whose sum of digits is also N
void smallestNumber( int N)
{
cout << (N % 9 + 1)
* pow (10, (N / 9))
- 1;
}

// Driver code
int main()
{
int N = 10;
smallestNumber(N);

return 0;
}

## Java

// Java program to find the smallest
// number whose sum of digits is also N
class GFG{

// Function to find the smallest
// number whose sum of digits is also N
static void smallestNumber( int N)
{
System.out.print((N % 9 + 1 ) *
Math.pow( 10 , (N / 9 )) - 1 );
}

// Driver code
public static void main(String[] args)
{
int N = 10 ;
smallestNumber(N);
}
}

// This code is contributed by sapnasingh4991

## Python3

# Python3 program to find the smallest
# number whose sum of digits is also N

# Function to find the smallest
# number whose sum of digits is also N
def smallestNumber(N):

print ((N % 9 + 1 ) * pow ( 10 , (N / / 9 )) - 1 )

# Driver code
N = 10
smallestNumber(N)

# This code is contributed by Code_Mech

## C#

// C# program to find the smallest
// number whose sum of digits is also N
using System;
class GFG{

// Function to find the smallest
// number whose sum of digits is also N
static void smallestNumber( int N)
{
Console.WriteLine((N % 9 + 1) *
Math.Pow(10, (N / 9)) - 1);
}

// Driver code
public static void Main()
{
int N = 10;

smallestNumber(N);
}
}

// This code is contributed by Ritik Bansal

19