hihoCoder 1234 Fractal——ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络赛

题目8 : Fractal

时间限制: 1000ms

单点时限: 1000ms

内存限制: 256MB

描述

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(1,1), 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, then square Ai+1Bi+1Ci+1Di+1 is generated by linking the midpoints of AiBi, BiCi, CiDi and DiAi successively.

4. Repeat step three 1000 times.

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

输入

In the first line there’s an integer T( T < 10,000), indicating the number of test cases.

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

输出

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

样例输入

3
0.375
0.001
0.478

样例输出

-1
4
20

/*********************************************************************/

题意：以A0(0,0), B0(0,1), C0(1,1), D0(1,0)四个点组成的正方形，每次取边A0B0, B0C0, C0D0, D0A0的中点相连，能得到新的正方形，按此操作1000次，可以得到如图的标识

现要在x为0~0.5之间作一条垂线x=k，问该垂线与标识有几个交点，当有无穷个交点时，直接输出"-1"

解题思路：高达90%正确率的题目不是水题也难呀，其实这题思路蛮简单的，首先，输出"-1"的情况无非就是垂线与小正方形的边重合；此外，越往内，每经过一个具有垂直于x轴的边的正方形，都会增加4个交点。

接下来就是暴力求解的工作了

每条垂直边的横坐标都等于(前一条垂直边的横坐标+0.5)/2

需要注意的一点是比赛的编译器不支持%I64d输入输出，仅支持%lld，比赛的时候还因为这个错了一次，虽然此题并不需要用到long long

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<deque>
#include<functional>
#include<iterator>
#include<set>
#include<utility>
#include<stack>
#include<queue>
using namespace std;
#define maxn 22222
#define inf 1500000007
int main()
{
    int i,j,k,n,m,T,p=1;
    int s;
    double x,y;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf",&x);
        y = 0.25;
        s = 4;
        if(x == 0)
        {
            printf("-1\n");
            continue;
        }
        while(1)
        {
            if(x == y)
            {
                s = -1;
                break;
            }
            if(x < y)
                 break;
            y = (y + 0.5)/2;
            s = s + 4;
        }
        printf("%d\n",s);
    }
    return 0;
}

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