CodeForces 233B Non-square Equation

http://blog.csdn.net/shuangde800/article/details/8082443

链接：

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

题目：

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.

分析与总结:

之前很少做数学题，所以一看到就想用二分法做，但是一直WA在test 5，后来经提醒可以将公式变形，瞬间明朗。

把公式x2 + s(x)·x - n = 0,
进行变形：

S(x) = n/x - x。

可大致估计S(x)的范围在1～100之间， 然后枚举S(x)的值，根据S(x)的值和方程S(x)
= n/x - x，解出x = sqrt( S(x)^2/4 + n ).
然后把x代入原公式x2 + s(x)·x - n = 0
看是否符合。

代码：

#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;
}