Dado un valor entero n, encuentre el n-ésimo entero positivo cuya suma es 10.
Ejemplos:
Input: n = 2 Output: 28 The first number with sum of digits as 10 is 19. Second number is 28. Input: 15 Output: 154
Método 1 (Simple):
Recorremos todos los números. Para cada número, encontramos la suma de los dígitos. Nos detenemos cuando encontramos el n-ésimo número con la suma de dígitos como 10.
C++
// Simple CPP program to find n-th number
// with sum of digits as 10.
#include <bits/stdc++.h>
using namespace std;
int findNth(int n)
{
int count = 0;
for (int curr = 1;; curr++) {
// Find sum of digits in current no.
int sum = 0;
for (int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
return -1;
}
int main()
{
printf("%d\n", findNth(5));
return 0;
}
Java
// Java program to find n-th number
// with sum of digits as 10.
import java.util.*;
import java.lang.*;
public class GFG {
public static int findNth(int n)
{
int count = 0;
for (int curr = 1;; curr++) {
// Find sum of digits in current no.
int sum = 0;
for (int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
}
public static void main(String[] args)
{
System.out.print(findNth(5));
}
}
// Contributed by _omg
Python3
# Python3 program to find n-th number # with sum of digits as 10. import itertools # function to find required number def findNth(n): count = 0 for curr in itertools.count(): # Find sum of digits in current no. sum = 0 x = curr while(x): sum = sum + x % 10 x = x // 10 # If sum is 10, we increment count if (sum == 10): count = count + 1 # If count becomes n, we return current # number. if (count == n): return curr return -1 # Driver program if __name__=='__main__': print(findNth(5)) # This code is contributed by # Sanjit_Prasad
C#
// C# program to find n-th number
// with sum of digits as 10.
using System;
class GFG {
public static int findNth(int n)
{
int count = 0;
for (int curr = 1;; curr++) {
// Find sum of digits in current no.
int sum = 0;
for (int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we
// return current number.
if (count == n)
return curr;
}
}
// Driver Code
static public void Main()
{
Console.WriteLine(findNth(5));
}
}
// This code is contributed
// by Sach_Code
PHP
<?php
// Simple PHP program to find n-th
// number with sum of digits as 10.
function findNth($n)
{
$count = 0;
for ($curr = 1; ; $curr++)
{
// Find sum of digits in
// current no.
$sum = 0;
for ($x = $curr;
$x > 0; $x = $x / 10)
$sum = $sum + $x % 10;
// If sum is 10, we increment
// count
if ($sum == 10)
$count++;
// If count becomes n, we return
// current number.
if ($count == $n)
return $curr;
}
return -1;
}
// Driver Code
echo findNth(5);
// This code is contributed by Sach .
?>
Javascript
<script>
// Simple JavaScript program to find n-th number
// with sum of digits as 10.
function findNth(n)
{
let count = 0;
for (let curr = 1;; curr++) {
// Find sum of digits in current no.
let sum = 0;
for (let x = curr; x > 0; x = Math.floor(x / 10))
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
return -1;
}
document.write(findNth(5));
// This code is contributed by Surbhi Tyagi.
</script>
Producción
55
Método 2 (Eficiente):
Si miramos más de cerca, podemos notar que todos los múltiplos de 9 están presentes en la progresión aritmética 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109,… Sin
embargo , hay números en la serie anterior cuya suma de dígitos no es 10, por ejemplo, 100. Entonces, en lugar de verificar uno por uno, comenzamos con 19 e incrementamos en 9.
C++
// Simple CPP program to find n-th number
// with sum of digits as 10.
#include <bits/stdc++.h>
using namespace std;
int findNth(int n)
{
int count = 0;
for (int curr = 19;; curr += 9) {
// Find sum of digits in current no.
int sum = 0;
for (int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
return -1;
}
int main()
{
printf("%d\n", findNth(5));
return 0;
}
Java
// Java program to find n-th number
// with sum of digits as 10.
import java.util.*;
import java.lang.*;
public class GFG {
public static int findNth(int n)
{
int count = 0;
for (int curr = 19;; curr += 9) {
// Find sum of digits in current no.
int sum = 0;
for (int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
}
public static void main(String[] args)
{
System.out.print(findNth(5));
}
}
// Contributed by _omg
Python3
# Python3 program to find n-th # number with sum of digits as 10. def findNth(n): count = 0; curr = 19; while (True): # Find sum of digits in # current no. sum = 0; x = curr; while (x > 0): sum = sum + x % 10; x = int(x / 10); # If sum is 10, we increment # count if (sum == 10): count+= 1; # If count becomes n, we return # current number. if (count == n): return curr; curr += 9; return -1; # Driver Code print(findNth(5)); # This code is contributed # by mits
C#
// C# program to find n-th number
// with sum of digits as 10.
using System;
class GFG {
public static int findNth(int n)
{
int count = 0;
for (int curr = 19;; curr += 9) {
// Find sum of digits in
// current no.
int sum = 0;
for (int x = curr;
x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment
// count
if (sum == 10)
count++;
// If count becomes n, we return
// current number.
if (count == n)
return curr;
}
}
// Driver Code
static public void Main()
{
Console.WriteLine(findNth(5));
}
}
// This code is contributed
// by Sach_Code
PHP
<?php
// Simple PHP program to find n-th
// number with sum of digits as 10.
function findNth($n)
{
$count = 0;
for ($curr = 19; ;$curr += 9)
{
// Find sum of digits in
// current no.
$sum = 0;
for ($x = $curr; $x > 0;
$x = (int)$x / 10)
$sum = $sum + $x % 10;
// If sum is 10, we increment
// count
if ($sum == 10)
$count++;
// If count becomes n, we return
// current number.
if ($count == $n)
return $curr;
}
return -1;
}
// Driver Code
echo findNth(5);
// This code is contributed
// by Sach_Code
?>
Javascript
<script>
// Simple Javascript program to find n-th
// number with sum of digits as 10.
function findNth(n)
{
let count = 0;
for (let curr = 19; ;curr += 9)
{
// Find sum of digits in
// current no.
let sum = 0;
for (let x = curr; x > 0;
x = parseInt(x / 10))
sum = sum + x % 10;
// If sum is 10, we increment
// count
if (sum == 10)
count++;
// If count becomes n, we return
// current number.
if (count == n)
return curr;
}
return -1;
}
// Driver Code
document.write(findNth(5));
// This code is contributed
// by gfgking
</script>
Producción
55
Publicación traducida automáticamente
Artículo escrito por Sumit bangar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA