求给定方程的解的个数
给定三个整数 A 、 B 和 C ，任务是求 X 的计数值，使得满足以下条件， X = B * Sm(X)A+C其中 Sm(X) 表示 X 和1<X<14 的位数之和 例:****
输入: A = 3，B = 2，C = 8 输出: 3 为 X = 10 ，2 * (1) 3 + 8 = 10 为 X = 2008 ，2 * (10) 3 + 8 = 2008 为 X = 13726
方法:这里可以做一个重要的观察，位数之和可以是 atmost 81(即 X = 999999999)，对应每个位数之和，我们得到一个单值 X 。所以我们可以迭代每个数字的和，检查 X 的结果值是否有效。 以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of
// valid values of X
int getCount(int a, int b, int c)
{
int count = 0;
// Iterate through all possible
// sum of digits of X
for (int i = 1; i <= 81; i++) {
// Get current value of X for sum of digits i
int cr = b * pow(i, a) + c;
int tmp = cr;
int sm = 0;
// Find sum of digits of cr
while (tmp) {
sm += tmp % 10;
tmp /= 10;
}
// If cr is a valid choice for X
if (sm == i && cr < 1e9)
count++;
}
// Return the count
return count;
}
// Driver code
int main()
{
int a = 3, b = 2, c = 8;
cout << getCount(a, b, c);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GfG
{
// Function to return the count of
// valid values of X
static int getCount(int a, int b, int c)
{
int count = 0;
// Iterate through all possible
// sum of digits of X
for (int i = 1; i <= 81; i++)
{
// Get current value of X for sum of digits i
int cr = b * (int)Math.pow(i, a) + c;
int tmp = cr;
int sm = 0;
// Find sum of digits of cr
while (tmp != 0)
{
sm += tmp % 10;
tmp /= 10;
}
// If cr is a valid choice for X
if (sm == i && cr < 1e9)
count++;
}
// Return the count
return count;
}
// Driver code
public static void main(String[] args)
{
int a = 3, b = 2, c = 8;
System.out.println(getCount(a, b, c));
}
}
// This code is contributed by Prerna Saini.
Python 3
# Python3 implementation of the approach
# Function to return the count of
# valid values of X
def getCount(a, b, c):
count = 0
# Iterate through all possible
# sum of digits of X
for i in range(1, 82):
# Get current value of X for
# sum of digits i
cr = b * pow(i, a) + c
tmp = cr
sm = 0
# Find sum of digits of cr
while (tmp):
sm += tmp % 10
tmp //= 10
# If cr is a valid choice for X
if (sm == i and cr < 10**9):
count += 1
# Return the count
return count
# Driver code
a, b, c = 3, 2, 8
print(getCount(a, b, c))
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count of
// valid values of X
static int getCount(int a, int b, int c)
{
int count = 0;
// Iterate through all possible
// sum of digits of X
for (int i = 1; i <= 81; i++)
{
// Get current value of X for sum
// of digits i
int cr = b * (int)Math.Pow(i, a) + c;
int tmp = cr;
int sm = 0;
// Find sum of digits of cr
while (tmp != 0)
{
sm += tmp % 10;
tmp /= 10;
}
// If cr is a valid choice for X
if (sm == i && cr < 1e9)
count++;
}
// Return the count
return count;
}
// Driver code
public static void Main()
{
int a = 3, b = 2, c = 8;
Console.Write(getCount(a, b, c));
}
}
// This code is contributed
// by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the count of
// valid values of X
function getCount($a, $b, $c)
{
$count = 0;
// Iterate through all possible
// sum of digits of X
for ($i = 1; $i <= 81; $i++)
{
// Get current value of X for sum of digits i
$cr = $b * (int)pow($i, $a) + $c;
$tmp = $cr;
$sm = 0;
// Find sum of digits of cr
while ($tmp != 0)
{
$sm += $tmp % 10;
$tmp /= 10;
}
// If cr is a valid choice for X
if ($sm == $i && $cr < 1e9)
$count++;
}
// Return the count
return $count;
}
// Driver code
{
$a = 3; $b = 2;$c = 8;
echo(getCount($a, $b, $c));
}
// This code is contributed by Code_Mech.
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to return the count of
// valid values of X
function getCount(a, b, c)
{
let count = 0;
// Iterate through all possible
// sum of digits of X
for (let i = 1; i <= 81; i++)
{
// Get current value of X for sum of digits i
let cr = b * Math.pow(i, a) + c;
let tmp = cr;
let sm = 0;
// Find sum of digits of cr
while (tmp != 0)
{
sm += tmp % 10;
tmp = Math.floor(tmp / 10);
}
// If cr is a valid choice for X
if (sm == i && cr < 1e9)
count++;
}
// Return the count
return count;
}
// driver program
let a = 3, b = 2, c = 8;
document.write(getCount(a, b, c));
</script>
Output:
3
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