给您一个整数n，找到等式x的最小正整数根，如果找不到根，则打印-1。
等式：x ^ 2 + s(x)* x – n = 0
其中x，n是正整数，s(x)是该函数，等于十进制数字系统中数字x的位数之和。
1 <= N <= 10 ^ 18
例子：
Input: N = 110
Output: 10
Explanation: x = 10 is the minimum root.
As s(10) = 1 + 0 = 1 and
102 + 1*10 - 110=0.
Input: N = 4
Output: -1
Explanation: there are no roots of the
equation possible
一个幼稚的方法是遍历X的所有可能值，并找出是否存在这样的根，但这将是不可能的，因为n的值非常大。
一种有效的方法如下
首先，让我们找出s(x)可能值的间隔。因此，x ^ 2 <= N并且N <= 10 ^ 18，x <=109。换句话说，对于每个相当大的解x，x的十进制长度都不会扩展10位。因此Smax = s(9999999999)= 10 * 9 = 90。
让我们蛮力s(x)的值(0 <= s(x)<= 90)。现在我们有了一个普通的平方方程。解决的办法是求解方程式，并检查s(x)的当前强力值等于解决方案的位数之和。如果解决方案存在并且等式成立，我们应该得到答案并存储尽可能少的根。
下面是上述方法的实现
C++
// CPP program to find smallest value of root
// of an equation under given constraints.
#include
using namespace std;
// function to check if the sum of digits is
// equal to the summation assumed
bool check(long long a, long long b)
{
long long int c = 0;
// calculate the sum of digit
while (a != 0) {
c = c + a % 10;
a = a / 10;
}
return (c == b);
}
// function to find the largest root possible.
long long root(long long n)
{
bool found = 0;
long long mx = 1e18;
// iterate for all possible sum of digits.
for (long long i = 0; i <= 90; i++) {
// check if discriminent is a perfect square.
long long s = i * i + 4 * n;
long long sq = sqrt(s);
// check if discriminent is a perfect square and
// if it as perefect root of the equation
if (sq * sq == s && check((sq - i) / 2, i)) {
found = 1;
mx = min(mx, (sq - i) / 2);
}
}
// function returns answer
if (found)
return mx;
else
return -1;
}
// driver program to check the above function
int main()
{
long long n = 110;
cout << root(n);
} Java
// Java program to find smallest value of root
// of an equation under given constraints.
class GFG{
// function to check if the sum of digits is
// equal to the summation assumed
static boolean check(long a, long b)
{
long c = 0;
// calculate the sum of digit
while (a != 0) {
c = c + a % 10;
a = a / 10;
}
return (c == b);
}
// function to find the largest root possible.
static long root(long n)
{
boolean found = false;
long mx = (long)1E18;
// iterate for all possible sum of digits.
for (long i = 0; i <= 90; i++) {
// check if discriminent is a perfect square.
long s = i * i + 4 * n;
long sq = (long)Math.sqrt(s);
// check if discriminent is a perfect square and
// if it as perefect root of the equation
if (sq * sq == s && check((sq - i) / 2, i)) {
found = true;
mx = Math.min(mx, (sq - i) / 2);
}
}
// function returns answer
if (found)
return mx;
else
return -1;
}
// driver program to check the above function
public static void main(String[] args)
{
long n = 110;
System.out.println(root(n));
}
}
// This code is contributed by mitsPython3
# Python3 program to find smallest
# value of root of an equation
# under given constraints.
import math
# function to check if the sum
# of digits is equal to the
# summation assumed
def check(a, b):
c = 0;
# calculate the
# sum of digit
while (a != 0):
c = c + a % 10;
a = int(a / 10);
return True if(c == b) else False;
# function to find the
# largest root possible.
def root(n):
found = False;
# float(1E+18)
mx = 1000000000000000001;
# iterate for all
# possible sum of digits.
for i in range(91):
# check if discriminent
# is a perfect square.
s = i * i + 4 * n;
sq = int(math.sqrt(s));
# check if discriminent is
# a perfect square and
# if it as perefect root
# of the equation
if (sq * sq == s and
check(int((sq - i) / 2), i)):
found = True;
mx = min(mx, int((sq-i) / 2));
# function returns answer
if (found):
return mx;
else:
return -1;
# Driver Code
n = 110;
print(root(n));
# This code is contributed by mitsC#
//C# program to find smallest value of root
// of an equation under given constraints.
using System;
public class GFG{
// function to check if the sum of digits is
// equal to the summation assumed
static bool check(long a, long b)
{
long c = 0;
// calculate the sum of digit
while (a != 0) {
c = c + a % 10;
a = a / 10;
}
return (c == b);
}
// function to find the largest root possible.
static long root(long n)
{
bool found = false;
long mx = (long)1E18;
// iterate for all possible sum of digits.
for (long i = 0; i <= 90; i++) {
// check if discriminent is a perfect square.
long s = i * i + 4 * n;
long sq = (long)Math.Sqrt(s);
// check if discriminent is a perfect square and
// if it as perefect root of the equation
if (sq * sq == s && check((sq - i) / 2, i)) {
found = true;
mx = Math.Min(mx, (sq - i) / 2);
}
}
// function returns answer
if (found)
return mx;
else
return -1;
}
// driver program to check the above function
public static void Main()
{
long n = 110;
Console.Write(root(n));
}
}
// This code is contributed by Raput-JiPHP
输出：
10