hdu 4342 History repeat itself 模拟题

题意：求从1开始的第n个非平方数以及图中公式

Historyrepeatitself

TimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)
TotalSubmission(s):875AcceptedSubmission(s):313

ProblemDescription

TomtooktheDiscreteMathematicscourseinthe2011,buthisbadattendanceangeredProfessorLeewhoisinchargeofthecourse.Therefore,ProfessorLeedecidedtoletTomfaceahardprobabilityproblem,andannouncedthatifhefailtoslovetheproblemtherewouldbenowayforTomtopassthefinalexam.
Asaresult,Tompassed.
Historyrepeatitself.You,thebadboy,alsoangeredtheProfessorLeewhenSeptemberEnds.Youhavetofacedtheproblemtoo.
TheproblemcomesthatYoumustfindtheN-thpositivenon-squarenumberMandprintedit.Andthat'sfornormalbadstudent,suchasTom.Buttherealbadstudenthastocalculatetheformulabelow.

So,thatyoucanreallyunderstandWHATABADSTUDENTYOUARE!!

Input

Thereisanumber(T)inthefirstline,tellyouthenumberoftestcasesbelow.ForthenextTlines,thereisjustonenumberontheeachlinewhichtellyoutheNofthecase.
Tosimplifiedtheproblem,TheNwillbewithin231andmorethen0.

Output

Foreachtestcase,printtheN-thnonsquarenumberandtheresultoftheformula.

SampleInput

4

1

3

6

10

SampleOutput

22

57

813

1328

/*假如让求第n个非平方数的话，看n前面有多少个平方数，假设有x个，则第n个非平方数就是n+x
则有 sqrt(n+x)=x  x>根号n     之后暴力   
对于第二个式子 由于下取整，所以每两个平方数之间的数的sqrt(i)，都等于前面的那个平方数，
 这样就很好计算了，一段一段的算
*/
#include <stdio.h>
#include <math.h>
int main()
{
	__int64 i,m,n,tt;
	__int64 x;
	__int64 s;
	while(scanf("%I64d",&m)!=EOF)
	{
		while(m--)
		{
			scanf("%I64d",&n);
			
			i=(__int64)sqrt(n*1.0);
			while((__int64)sqrt((n+i)*1.0)!=i)//这里必须要乘1.0 否则编译错误
			{
				i++;
			}
			x=n+i;
			n=n+i;
			tt=(__int64)(sqrt(x*1.0));
			s=0;
			for(i=2;i<=tt;i++)
			{
				s=s+(i*i-(i-1)*(i-1))*(i-1);//从i*i到 (i-1)*(i-1)的数  开方后向下取整 均为i-1
			}
			s=s+(x-tt*tt+1)*tt;
			printf("%I64d %I64d\n",n,s);
		}
	}
	return 0;
}