找到一个数字 x，使得 x 和它的数字之和等于给定的 n。
原文:https://www . geesforgeks . org/find-number-x-sum-x-digits-equal-given-n/
给定一个正数 n .我们需要找到一个数 x，使得 x 对其本身的位数之和等于 n. 如果没有这样的 x 是可能的，打印-1。 例:
Input : n = 21
Output : x = 15
Explanation : x + its digit sum = 15 + 1 + 5 = 21
Input : n = 5
Output : -1
我们从 1 迭代到 n，对于每个中间数，x 找到它的数字和，然后加到 x 上，如果等于 n，那么 x 就是我们需要的答案。
// iterate from 1 to n. For every no.
// check if its digit sum with it is
// equal to n.
for (int i = 0; i <= n; i++)
if (i + digSum(i) == n)
return i;
return -1;
C++
// CPP program to find x such that x +
// digSum(x) is equal to n.
#include <bits/stdc++.h>
using namespace std;
// utility function for digit sum
int digSum(int n)
{
int sum = 0, rem = 0;
while (n) {
rem = n % 10;
sum += rem;
n /= 10;
}
return sum;
}
// function for finding x
int findX(int n)
{
// iterate from 1 to n. For every no.
// check if its digit sum with it is
// equal to n.
for (int i = 0; i <= n; i++)
if (i + digSum(i) == n)
return i;
// if no such i found return -1
return -1;
}
// driver function
int main()
{
int n = 43;
cout << "x = " << findX(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find x such that x +
// digSum(x) is equal to n.
class GFG
{
// utility function for digit sum
static int digSum(int n)
{
int sum = 0, rem = 0;
while (n>0) {
rem = n % 10;
sum += rem;
n /= 10;
}
return sum;
}
// function for finding x
static int findX(int n)
{
// iterate from 1 to n. For every no.
// check if its digit sum with it is
// equal to n.
for (int i = 0; i <= n; i++)
if (i + digSum(i) == n)
return i;
// if no such i found return -1
return -1;
}
// Driver code
public static void main (String[] args)
{
int n = 43;
System.out.println("x = "+findX(n));
}
}
// This code is contributed by Anant Agarwal.
Python 3
# Python3 program to find
# x such that dx + igSum(x)
# is equal to n.
# utility function
# for digit sum
def digSum(n):
sum = 0;
rem = 0;
while(n):
rem = n % 10;
sum = sum + rem;
n = int(n / 10);
return sum;
# function for finding x
def findX(n):
# iterate from 1 to n.
# For every no.
# check if its digit
# sum with it is# equal to n.
for i in range(n + 1):
if (i + digSum(i) == n):
return i;
# if no such i
# found return -1
return -1;
# Driver Code
n = 43;
print("x = ", findX(n));
# This code is contributed by mits
C
// C# program to find x such that
// x + digSum(x) is equal to n.
using System;
class GFG {
// utility function for digit sum
static int digSum(int n)
{
int sum = 0, rem = 0;
while (n > 0)
{
rem = n % 10;
sum += rem;
n /= 10;
}
return sum;
}
// function for finding x
static int findX(int n)
{
// iterate from 1 to n. For every no.
// check if its digit sum with it is
// equal to n.
for (int i = 0; i <= n; i++)
if (i + digSum(i) == n)
return i;
// if no such i found return -1
return -1;
}
// Driver code
public static void Main()
{
int n = 43;
Console.Write("x = " + findX(n));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find x such that
// dx + igSum(x) is equal to n.
// utility function
// for digit sum
function digSum($n)
{
$sum = 0; $rem = 0;
while ($n)
{
$rem = $n % 10;
$sum += $rem;
$n /= 10;
}
return $sum;
}
// function for finding x
function findX($n)
{
// iterate from 1 to n.
// For every no.
// check if its digit
// sum with it is
// equal to n.
for ($i = 0; $i <= $n; $i++)
if ($i + digSum($i) == $n)
return $i;
// if no such i
// found return -1
return -1;
}
// Driver Code
$n = 43;
echo "x = " , findX($n);
// This code is contributed by vt_m.
?>
java 描述语言
<script>
// Javascript program to find x such that x +
// digSum(x) is equal to n.
// utility function for digit sum
function digSum(n)
{
let sum = 0, rem = 0;
while (n>0) {
rem = n % 10;
sum += rem;
n = Math.floor(n / 10);
}
return sum;
}
// function for finding x
function findX(n)
{
// iterate from 1 to n. For every no.
// check if its digit sum with it is
// equal to n.
for (let i = 0; i <= n; i++)
if (i + digSum(i) == n)
return i;
// if no such i found return -1
return -1;
}
// driver program
let n = 43;
document.write("x = "+findX(n));
</script>
输出:
x = 35
本文由Shivam Pradhan(anuj _ charm)供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
版权属于：月萌API www.moonapi.com，转载请注明出处
