codeforces-233【B思维】

题目链接：点击打开链接

B. Non-square Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Let's consider equation:

x2 + s(x)·x - n = 0, 

where x, n are positive integers, s(x) is the function, equal to the sum of digits of number x in the decimal number system.

You are given an integer n, find the smallest positive integer root of equation x, or else determine that there are no such roots.

Input

A single line contains integer n (1 ≤ n ≤ 1018) — the equation parameter.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64dspecifier.

Output

Print -1, if the equation doesn't have integer positive roots. Otherwise print such smallest integer x (x > 0), that the equation given in the statement holds.

Examples input

2
      

output

1
      

input

110
      

output

10
      

input

4
      

output

-1
      

Note

In the first test case x = 1 is the minimum root. As s(1) = 1 and 12 + 1·1 - 2 = 0.

In the second test case x = 10 is the minimum root. As s(10) = 1 + 0 = 1 and 102 + 1·10 - 110 = 0.

In the third test case the equation has no roots.

大意：给你一个方程，让你找到一个方程的正整数根，其中 n 为一个参数，s(x) 为十进制下根 x 的每一位数字上之和

思路：这里给出的 n 的取值为 1e18，令方程变为 x * x + s(x) * x = n 则可以看出，根 x 的取值最大不会超过 1e9，也就是 x 的每一位数字之和加起来 9*9 = 81，就是 s(x) 的值不会超过 81，然后枚举 s(x)，找到 x 即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
LL n;
LL dig_sum(LL x)
{
	LL ans=0;
	while(x)
	{
		ans+=x%10;
		x/=10;
	}
	return ans;
}
double solve(LL x,LL n) // 求根公式，题中说了要正数解，所以这里这里只考虑正数解 
{
	return (sqrt(x*x+4*n+0.0)-x*1.0)/2;
}
int main()
{
	while(~scanf("%I64d",&n))
	{
		LL ans=-1;
		for(LL i=1;i<=90;i++)
		{
			LL root=1LL*solve(i,n);
			LL sum=dig_sum(root);
			if(root*root+sum*root-n==0)
			{
				ans=root;
				break;
			}
		}
		printf("%I64d\n",ans);
	}
	return 0;
}