Hihocoder #1234: Fractal (Math simple problem)

Time limit: 1000ms single point time limit: 1000ms memory limit: 256MB description

This is the logo of PKUACM 2016. More specifically, the logo is generated as follows:

1. Put Four Points A0 (0,0), B0 (0,1), C0 (All), D0 (1,0) on a Cartesian coordinate system.

2. Link a0b0, b0c0, c0d0, d0a0 separately, forming square a0b0c0d0.

3. Assume we have already generated square Aibicidi and then square ai+1bi+1ci+1di+1 are generated by linking the midpoints of Aibi, Bici, Cidi and Diai successively.

4. Repeat step three.

Now the designer decides-add a vertical line x=k-to-this logo (0<= K < 0.5, and for K, there'll be is at most 8 di Gits after the decimal point). He wants to know the number of common points between the new line and the original logo.

Input

In the first line there's an integer T (T < s), indicating the number of test cases.

Then T-lines follow, each describing a-test case. Each line contains a float number k, meaning that is should calculate the number of common points between line x = K and The logo.

Output

For each test case, print a line containing one integer indicating the answer. If There is infinity common points, print-1.

Sample input

30.3750.0010.478

Sample output

-1420


Find out the rules of calculation

1 #pragmaComment (linker, "/stack:1024000000,1024000000")2#include <iostream>3#include <cstdio>4#include <cstring>5#include <cmath>6#include <math.h>7#include <algorithm>8#include <queue>9#include <Set>Ten#include <bitset> One#include <map> A#include <vector> -#include <stdlib.h> -#include <stack> the using namespacestd; - intdirx[]={0,0,-1,1}; - intdiry[]={-1,1,0,0}; - #definePI ACOs (-1.0) + #defineMax (a) (a) > (b)? (a): (b) - #defineMin (a) (a) < (b)? (a): (b) + #definell Long Long A #defineEPS 1e-10 at #defineMOD 1000000007 - #defineN 1000000 - #defineINF 1e12 - DoubleN; - intnum[100000]; -  in intMain () - { to     intT; +scanf"%d",&t); -      while(t--){ thescanf"%LF",&n); *         inti; $         Doublex=0;Panax Notoginseng         Doubley=0.5; -          for(i=0;; i++){ the             if(x>=N) { +                  Break; A             } thex=x+y/2.0; +Y/=2.0; -         } $         intw=i; $         if(x==N) { -printf"-1\n"); -         } the         Else{ -             intans=0;Wuyi              for(intI=0; i<w;i++){ theans+=4; -             } Wuprintf"%d\n", ans); -         } About     } $     return 0; -}

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Hihocoder #1234: Fractal (Math simple problem)