Non-square Equation【公式轉換】

Non-square Equation

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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.

【題意】根據公式，給出n，求能滿足公式的最小正整數，S(x) 爲x每一位數字的和， 不存在x滿足公式時，輸出-1。

【思路】公式轉換

   s(x) 函數，可以發現 s(x) 在 x 取得最大值，即 x = 999999999 時相應地取得最大值，即 81。於是，我們可以去枚舉 s(x) ，計算相應的 x 能否使方程左右兩邊相等，這樣就能大大降低複雜度。利用配方公式，可以得到x = sqrt( S(x)^2 /4 + n ) - S(x)/2。如此一來，我們只需要從 1 到 81 枚舉所有 s(x)，然後找到對應的最小x，代入方程檢查是否相等符合。

AC 代碼：

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

int get(__int64 x) 
{
    int sum = 0;
    while(x) 
	{
        sum += x%10;
        x /= 10;
    }
    return sum;
}
int main() 
{
    __int64 n, ans = -1;
    scanf("%I64d", &n);
    for(int i=1; i<=81; ++i) 
	{
        __int64 x = sqrt(i*i/4+n) - i/2;
        int sx = get(x);
        if(x*x+sx*x-n == 0) 
		{
            ans = x;
            break;
        }
    }
    printf("%I64d\n", ans);
    return 0;
}