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.
解题思路：
原方程变形后：s(x)=n/x-x;可以推测s(x)的范围为1-100.然后利用s(x)求解x。最后将s(x)和x带入原方程。若成立，则x为所求。

#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;
}