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
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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;
}
作者：csdn博客 shuangde800
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