題目：
B. Non-square Equation
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 %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 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.
分析與總結:
之前很少做數學題，所以一看到就想用二分法做，但是一直WA在test 5，後來經提醒可以將公式變形，瞬間明朗。
把公式x2 + s(x)·x - n = 0, 進行變形：
S(x) = n/x - x。
可大致估計S(x)的范圍在1～100之間， 然後枚舉S(x)的值，根據S(x)的值和方程S(x) = n/x - x，解出x = sqrt( S(x)^2/4 + n ).
然後把x代入原公式x2 + s(x)·x - n = 0 看是否符合。
代碼：
[cpp]
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long int64;
int64 n, sx;
int64 digitSum(int64 n){
int64 sum=0;
while(n){
sum += n%10;
n/=10;
}
return sum;
}
int main(){
while(cin >> n){
int64 x=1, end=1e8, ans=-1;
for(int64 i=1; i<=100; ++i){
int64 tmp = i*i/4+n;
x = sqrt(tmp)-i/2;
sx = digitSum(x);
if(x*x+sx*x-n==0){
ans=x;
break;
}
}
cout << ans << endl;
}
return 0;
}
