CodeForces-233B-Non-square Equation

Description
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 %I64d specifier.

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.

Sample Input
Input
2
Output
1
Input
110
Output
10
Input
4
Output
-1
Hint
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.
解題思路：
原方程變形後：s(x)=n/x-x;可以推測s(x)的範圍爲1-100.然後利用s(x)求解x。最後將s(x)和x帶入原方程。若成立，則x爲所求。

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;


typedef long long LL;


LL digitSum(LL n)
{
    LL sum = 0;
    while(n)
    {
        sum+=n%10;
        n/=10;
    }
    return sum;
}


LL n;

int main()
{
    while(cin >> n)
    {

        LL ds,ans=-1,x,tmp;

        for ( int i = 1; i <= 100; ++i)
        {
            tmp = i*i/4+n;
            x= -i/2+sqrt(tmp);
            ds=digitSum(x);
            if (x*x+ds*x-n==0){
                ans=x;
                break;
            }
        }
        cout<<ans<<endl;

    }
    return 0;
}