题意：给n 求最小的x
题解：二元一次方程求解，x*x+x*b-n=0 遍历b
Description
Let's consider equation:
x^2 + 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.
1 #include<stdio.h> 2 #include<math.h> 3 4 __int64 find(__int64 x) 5 { 6 __int64 b=0; 7 while(x) 8 { 9 b+=x%10; 10 x=x/10; 11 } 12 return b; 13 } 14 15 int main() 16 { 17 __int64 n,i,d,x; 18 while(~scanf("%I64d",&n)) 19 { 20 __int64 ans=-1; 21 for(i=1;i<=100;i++) 22 { 23 d=sqrt((double)(i*i+4*n)); 24 x=d-i; 25 if(x%2==0&&find(x/2)==i) 26 { 27 ans=x/2; 28 break; 29 } 30 } 31 printf("%I64d ",ans); 32 } 33 34 return 0; 35 }