求前 N 个整数，使其位数之和等于 10
原文:https://www . geesforgeks . org/find-first-n-integer-so-它们的数字总和等于-10/
给定一个整数 N ，任务是先打印数字之和为 10 的 N 个整数。 举例:
输入: N = 4 输出: 19 28 37 46 输入: N = 6 输出: 19 28 37 46 55 64
方法:初始化 num = 19 得到该系列的第一个号码，现在将 9 加到前面的号码上，检查新号码的位数之和是否为 10 。如果是，则这是该系列的下一个编号，这是因为所需系列的任意两个连续编号之间的差值至少为9。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// sum of digits of n
int sum(int n)
{
int sum = 0;
while (n) {
// Add the last digit to the sum
sum = sum + n % 10;
// Remove last digit
n = n / 10;
}
// Return the sum of digits
return sum;
}
// Function to print the first n numbers
// whose sum of digits is 10
void firstN(int n)
{
// First number of the series is 19
int num = 19, cnt = 1;
while (cnt != n) {
// If the sum of digits of the
// current number is equal to 10
if (sum(num) == 10) {
// Print the number
cout << num << " ";
cnt++;
}
// Add 9 to the previous number
num += 9;
}
}
// Driver code
int main()
{
int n = 10;
firstN(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.util.*;
class GFG
{
// Function to return the
// sum of digits of n
static int sum(int n)
{
int sum = 0;
while (n > 0)
{
// Add the last digit to the sum
sum = sum + n % 10;
// Remove last digit
n = n / 10;
}
// Return the sum of digits
return sum;
}
// Function to print the first n numbers
// whose sum of digits is 10
static void firstN(int n)
{
// First number of the series is 19
int num = 19, cnt = 1;
while (cnt != n)
{
// If the sum of digits of the
// current number is equal to 10
if (sum(num) == 10)
{
// Print the number
System.out.print(num + " ");
cnt++;
}
// Add 9 to the previous number
num += 9;
}
}
// Driver code
public static void main(String[] args)
{
int n = 10;
firstN(n);
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
# Function to return the
# sum of digits of n
def sum(n) :
sum = 0;
while (n) :
# Add the last digit to the sum
sum = sum + n % 10;
# Remove last digit
n = n // 10;
# Return the sum of digits
return sum;
# Function to print the first n numbers
# whose sum of digits is 10
def firstN(n) :
# First number of the series is 19
num = 19; cnt = 1;
while (cnt != n) :
# If the sum of digits of the
# current number is equal to 10
if (sum(num) == 10) :
# Print the number
print(num,end= " ");
cnt += 1;
# Add 9 to the previous number
num += 9;
# Driver code
if __name__ == "__main__" :
n = 10;
firstN(n);
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the
// sum of digits of n
static int sum(int n)
{
int sum = 0;
while (n > 0)
{
// Add the last digit to the sum
sum = sum + n % 10;
// Remove last digit
n = n / 10;
}
// Return the sum of digits
return sum;
}
// Function to print the first n numbers
// whose sum of digits is 10
static void firstN(int n)
{
// First number of the series is 19
int num = 19, cnt = 1;
while (cnt != n)
{
// If the sum of digits of the
// current number is equal to 10
if (sum(num) == 10)
{
// Print the number
Console.Write(num + " ");
cnt++;
}
// Add 9 to the previous number
num += 9;
}
}
// Driver code
public static void Main(String[] args)
{
int n = 10;
firstN(n);
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the
// sum of digits of n
function sum(n) {
var sum = 0;
while (n > 0) {
// Add the last digit to the sum
sum = sum + n % 10;
// Remove last digit
n = parseInt(n / 10);
}
// Return the sum of digits
return sum;
}
// Function to print the first n numbers
// whose sum of digits is 10
function firstN(n) {
// First number of the series is 19
var num = 19, cnt = 1;
while (cnt != n) {
// If the sum of digits of the
// current number is equal to 10
if (sum(num) == 10) {
// Print the number
document.write(num + " ");
cnt++;
}
// Add 9 to the previous number
num += 9;
}
}
// Driver code
var n = 10;
firstN(n);
// This code contributed by Rajput-Ji
</script>
Output:
19 28 37 46 55 64 73 82 91
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