Codeforces 233B Non-square equation

Link:

Http://codeforces.com/problemset/problem/233/B

Topic:

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 a integer n, find the smallest positive integer root of equation x, or else determine that there are no suc H roots.
Input

A single line contains integer n (1≤n≤1018)-the equation parameter.

Please, don't use the%lld specifier to read or write 64-bit integers inс++. It is preferred to use CIN, cout streams or the%i64dspecifier.
Output

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Print-1, if the equation doesn ' t have integer positive roots. Otherwise print such smallest integer x (x>0), which is equation given in the statement.
Sample Test (s)
Input

2

Output

1

Input

110

Output

10

Input

4

Output

-1

Note

In the ' The ' the ' 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 (=1+0=1) and 102+1 10-110=0.

In the third test case the equation has no roots.

Analysis and Summary:

Before the math problem, so the first to see the use of the two-way, but has been WA in Test 5, and then the reminder can be distorted formula, instantaneous clarity.

x-n=0 The formula X2+s (x) to deform:

S (x) = n/x-X.

You can roughly estimate the range of s (x) between 1~100, and then enumerate the values of s (x), and then, according to the value of s (x) and the equation s (x) = n/x-X, solve the x = sqrt (s (x) ^2/4 + N).

The x is then x2+s to the original formula (x) x-n=0 to see if it fits.

Code:

#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;  
}

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