[思维]Folding a Ribbon

题目描述

Think of repetitively folding a very long and thin ribbon. First, the ribbon is spread out from left to right, then it is creased at its center, and one half of the ribbon is laid over the other. You can either fold it from the left to the right, picking up the left end of the ribbon and laying it over the right end, or from the right to the left, doing the same in the reverse direction. To fold the already folded ribbon, the whole layers of the ribbon are treated as one thicker ribbon, again from the left to the right or the reverse. 
After folding the ribbon a number of times, one of the layers of the ribbon is marked, and then the ribbon is completely unfolded restoring the original state. Many creases remain on the unfolded ribbon, and one certain part of the ribbon between two creases or a ribbon end should be found marked. Knowing which layer is marked and the position of the marked part when the ribbon is spread out, can you tell all the directions of the repeated folding, from the left or from the right? 
The figure below depicts the case of the first dataset of the sample input. 

 

输入

The input consists of at most 100 datasets, each being a line containing three integers. 
n i j
The three integers mean the following: The ribbon is folded n times in a certain order; then, the i-th layer of the folded ribbon, counted from the top, is marked; when the ribbon is unfolded completely restoring the original state, the marked part is the j-th part of the ribbon separated by creases, counted from the left. Both i and j are one-based, that is, the topmost layer is the layer 1 and the leftmost part is numbered 1. These integers satisfy 1 ≤ n ≤ 60, 1 ≤ i ≤ 2n, and 1 ≤ j ≤ 2n. 
The end of the input is indicated by a line with three zeros. 
 

 

输出

For each dataset, output one of the possible folding sequences that bring about the result specified in the dataset. 
The folding sequence should be given in one line consisting of n characters, each being either L or R. L means a folding from the left to the right, and R means from the right to the left. The folding operations are to be carried out in the order specified in the sequence. 
 

 

样例输入

3 3 2
12 578 2214
59 471605241352156968 431565444592236940
0 0 0

 

样例输出

LRR
RLLLRRRLRRLL
LRRRLRRLLRRRRLLLLRLLRRRLRRLLRLLLLLLRLRLLRLRLLLRLRLLRLLRRRLL

 

代码如下:

 1 #include <iostream>
 2 #include <bits/stdc++.h>
 3 using namespace std;
 4 typedef long long ll;
 5 ll n,x,y;
 6 ll pw[100];
 7 ll tot[150];
 8 void init()
 9 {
10     pw[0]=1;
11     for(int i=1;i<=62;i++)
12     {
13         pw[i]=pw[i-1]<<1;
14     }
15 }
16 void solve(){
17     for(int i=1;i<=n;i++)
18     {
19         //cout << y <<" " <<pw[n-i] << endl;
20         if(y<=pw[n-i])
21         {
22             if(tot[i]<=pw[i]/2)
23             {
24                 cout << "R";
25  
26             }
27             else
28             {
29                 cout << "L";
30                 y=pw[n-i]+1-y;
31  
32             }
33         }
34         else
35         {
36             if(tot[i]>pw[i]/2)
37             {
38                 cout << "R";
39                 y=pw[n-i+1]-y+1;
40             }
41             else
42             {
43                 cout << "L";
44                 y=y-pw[n-i];
45             }
46         }
47     }
48     cout << endl;
49 }
50 int main()
51 {
52     init();
53     while(cin>>n>>x>>y)
54     {
55         if(n+x+y==0)
56             break;
57         ll down=pw[n]+1-x;
58         tot[n]=down;
59         for(int i=n;i>1;i--)
60         {
61             if(down>pw[i]/2) //可以写成pow_2[i-1]
62             {
63                 down=pw[i]-down+1;
64             }
65             tot[i-1]=down;
66         }
67         solve();
68     }
69     //cout << "Hello world!" << endl;
70     return 0;
71 }
View Code

参考博客:https://blog.csdn.net/lingyan_BVB/article/details/82317739

老赵tql

posted @ 2018-09-03 20:20  听风不成泣  阅读(247)  评论(0编辑  收藏  举报