• 树的 起步*------二叉树


    A number of K balls are dropped one by one from the root of a fully binary tree structure FBT. Each
    time the ball being dropped first visits a non-terminal node. It then keeps moving down, either follows
    the path of the left subtree, or follows the path of the right subtree, until it stops at one of the leaf
    nodes of FBT. To determine a ball’s moving direction a flag is set up in every non-terminal node with
    two values, either false or true. Initially, all of the flags are false. When visiting a non-terminal node
    if the flag’s current value at this node is false, then the ball will first switch this flag’s value, i.e., from
    the false to the true, and then follow the left subtree of this node to keep moving down. Otherwise,
    it will also switch this flag’s value, i.e., from the true to the false, but will follow the right subtree of
    this node to keep moving down. Furthermore, all nodes of FBT are sequentially numbered, starting at
    1 with nodes on depth 1, and then those on depth 2, and so on. Nodes on any depth are numbered
    from left to right.
    For example, Fig. 1 represents a fully binary tree of maximum depth 4 with the node numbers 1,
    2, 3, ..., 15. Since all of the flags are initially set to be false, the first ball being dropped will switch
    flag’s values at node 1, node 2, and node 4 before it finally stops at position 8. The second ball being
    dropped will switch flag’s values at node 1, node 3, and node 6, and stop at position 12. Obviously,
    the third ball being dropped will switch flag’s values at node 1, node 2, and node 5 before it stops at
    position 10.
    Fig. 1: An example of FBT with the maximum depth 4 and sequential node numbers.
    Now consider a number of test cases where two values will be given for each test. The first value is
    D, the maximum depth of FBT, and the second one is I, the I-th ball being dropped. You may assume
    the value of I will not exceed the total number of leaf nodes for the given FBT.
    Please write a program to determine the stop position P for each test case.
    For each test cases the range of two parameters D and I is as below:
    2 ≤ D ≤ 20, and 1 ≤ I ≤ 524288.
    Input
    Contains l + 2 lines.
    Line 1 l the number of test cases
    Line 2 D1 I1 test case #1, two decimal numbers that are separated by one blank
    ...
    Line k + 1 Dk Ik test case #k
    Line l + 1 Dl Il test case #l
    Line l + 2 -1 a constant ‘-1’ representing the end of the input file
    Output
    Contains l lines.
    Line 1 the stop position P for the test case #1
    ...
    Line k the stop position P for the test case #k
    ...
    Line l the stop position P for the test case #l
    Sample Input
    5
    4 2
    3 4
    10 1
    2 2
    8 128
    -1
    Sample Output
    12
    7
    512
    3
    255

    给你一个完全二叉树 其深度为D  从上到下 从左到右的  叶子 分别 编号为   1  -  (2^D) -1        每个叶子上  都有一个开关  起始的时候   开关都是 关闭的      当开关关闭的 时候   小球都是 向 该节点的 左儿子 走去开关开启的  时候 相反  

    现在 给你 树的深度 D  和  小球的  个数   I  问你   最后 一个 小球  落在了 那个  叶子节点上 .

     1 #include<cstdio>
     2 #include<cstring>
     3 const int maxd=20 ;
     4 int s[1<<maxd];
     5 int main()
     6 {
     7     int D,I;
     8     while(scanf("%d%d",&D,&I)==2)
     9     {
    10         memset(s,0,sizeof(s));             //  表示 开关的 状态 .
    11         int k,n=(1<<D)-1;        //  编号最大的 叶子
    12         for(int i=1;i<=I;i++)
    13         {
    14             k=1;
    15             for(;;)
    16             {
    17                 s[k]=!s[k];
    18                 k=s[k]?2*k:k*2+1;
    19                 if(k>n)
    20                     break;
    21             }
    22         }
    23         printf("%d",k/2);
    24     }
    25     return 0;
    26 }
  • 相关阅读:
    codeforces C. Cows and Sequence 解题报告
    codeforces A. Point on Spiral 解题报告
    codeforces C. New Year Ratings Change 解题报告
    codeforces A. Fox and Box Accumulation 解题报告
    codeforces B. Multitasking 解题报告
    git命令使用
    shell简单使用
    知识束缚
    php 调用系统命令
    数据传输方式(前端与后台 ,后台与后台)
  • 原文地址:https://www.cnblogs.com/A-FM/p/5247410.html
Copyright © 2020-2023  润新知