Hihocoder 1234 Fractal (Law of Search)

Topic Links:

http://hihocoder.com/problemset/problem/1234

Problem Solving Ideas:

Main topic:

With A0 (0,0), B0 (0,1), C0 (total), D0 (1,0) Four points of the square, each take the edge a0b0, B0C0, c0d0, d0a0 midpoint connected, can get a new square, press

This operation 1000 times, you can get with the title given to the picture, now in X for 0~0.5 between a vertical x=k, ask you the vertical line and the logo has a few intersections, when there are infinite

Output "-1" directly at the intersection of

Algorithm idea:

A closer look will reveal that the horizontal axis of each vertical edge is equal to (the horizontal axis of the previous vertical edge +0.5)/2, and then according to: if the vertical line is coincident with the edge of the small square,

The output is "1"; the other, the farther inward, each passing through a square with an edge perpendicular to the x-axis, adds 4 intersections. Finally, it can be solved by lower_bound ...

AC Code:

#include <iostream>
#include <cstdio>
using namespace std;

Double a[1005];

void Init () {
    double x = 0.5;
    for (int i = 0; I <=, i++) {
        a[i] = 0.5-x;
        x *= 0.5;
    }
}

int main () {
    int T;
    Init ();
    scanf ("%d", &t);
    while (t--) {
        double K;
        scanf ("%lf", &k);
        int pos = Lower_bound (a,a+505,k)-A;
        if (a[pos] = = k)
            puts ("-1");
        else
            printf ("%d\n", 4*pos);
    }
    return 0;
}