CodeForces 233B Non-square Equation （数学）

A - Non-square Equation
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

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 %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.

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.

如果枚举x的话，数字范围太大，所以枚举S(x),解方程求x;

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
__int64 init(__int64 x)
{
__int64 sum=0;
while(x)
{
sum+=x%10;
x=x/10;
}
return sum;
}
int main()
{
__int64 n,i,j,k,l,b,m;
while(scanf("%I64d",&n)!=EOF)
{
int flag=0;
for(i=1;i<=100;i++)
{
m=sqrt((double)(i*i+4*n));
b=m-i;
if(b%2==0&&(init(b/2))==i)
{
flag=1;
break;
}
}
if(flag==0)
printf("-1\n");
else
{
printf("%I64d\n",b/2);
}
}
return 0;
}