CodeForces 233B. Non-square Equation【狀壓】

DTL66發表於2016-11-07

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 %I64d specifier.

Output

Print -1, if the equation doesn't have integer positive roots. Otherwise print such smallest integerx (x > 0), that the equation given in the statement holds.

Examples
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 and102 + 1·10 - 110 = 0.

In the third test case the equation has no roots.


AC程式碼： 

#include<cstdio>
#include<cmath>

typedef __int64 LL;

bool Judge(LL x,LL s) {
	LL sum=0;
	while(x) {
		sum=sum+x%10; x/=10;
	} 
	return sum==s; 
}
int main()
{
	LL N;
	while(~scanf("%I64d",&N)) {
        LL n=17*9,ans=-1;
		for(int i=1;i<=n;++i) {
		    LL t=i*i+4*N;
			LL tem=sqrt(t);
			if(tem*tem==t)  if( (tem-i)%2==0) {
				 	LL x=(tem-i)>>1;
				    if(Judge(x,i)) {
				    	ans=x; break;
					}
			}
		} 
		printf("%I64d\n",ans); 
	}
	return 0;
 }


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