Codeforces 233B Non-square Equation (数学+思维)

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%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 test(s)

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.

解题说明：刚开始想的是二分，最后还是不对，这是道数学题，因为n的最大值为1018，所以x的最大值为109 ，s(x)取最大值时x=999999999，也就是s(x)最大值为81，只需从1-81列举一下就行。

计算相应的 x 能否使方程左右两边相等，这样就能大大降低复杂度。我们将原方程化为左边是 x 的方程：

x2+s(x)⋅x−n+s(x)24=s(x)24(1)

(x+s(x)2)2=s(x)24+n(2)

x+s(x)2=s(x)24+n−−−−−−−√(3)

x=s(x)24+n−−−−−−−√−s(x)2(4)

这样，我们只需要从 1 到 81 枚举所有 s(x)，计算出相应的 x，代入方程比较是否相等即可。

#include<cstdio>
#include<cmath>
#define LL long long
int main()
{
LL n,i,s,x,t;
scanf("%I64d",&n);
for(i=1;i<=81;i++)
{
s=0;
x=sqrt(i*i/4+n)-i/2;
t=x;
while(t)
{
s+=t%10;
t=t/10;
}
if(x*x+s*x-n==0)
{
printf("%I64d\n",x);
break;
}
}
if(i==82)
printf("-1\n");
return 0;
} </span>