Dropping Balls (二叉树+思维)
2015-12-28 20:53
387 查看
Dropping Balls |
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 befalse, 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 Ith 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:
Input
Contains l+2 lines.Line 1 I the number of test cases Line 2 test case #1, two decimal numbers that are separatedby one blank ... Line k+1 test case #k Line l+1 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
题解:一个小球从二叉树的顶端开始往下落,刚开始二叉树结点开关闭合,如果闭合向左树落,否则右树,问第I个球会落到哪个结点;递归超时;
当然也可以把球数看为切入点,每次球往下落,落到此结点的次数是上一个节点次数的一半;
ac代码:
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<stack> #include<vector> using namespace std; const int INF=0x3f3f3f3f; const double PI=acos(-1.0); #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define SL(x) scanf("%lld",&x) #define PI(x) printf("%d",x) #define PL(x) printf("%lld",x) #define P_ printf(" "); #define T_T while(T--) int main(){ int D,I; int T; while(SI(T),T!=-1){ T_T{ SI(D);SI(I); int cur=1,root=1; while(++cur<=D){ if(I%2==1){ root=root<<1; } else root=root<<1|1; I=(I+1)/2; } printf("%d\n",root); } } return 0; }
超时代码:因为球的数目可能很多,对于每一个暴力,肯定要超时了
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<stack> #include<vector> using namespace std; const int INF=0x3f3f3f3f; const double PI=acos(-1.0); #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define SL(x) scanf("%lld",&x) #define PI(x) printf("%d",x) #define PL(x) printf("%lld",x) #define P_ printf(" "); #define T_T while(T--) #define lson root<<1 #define rson root<<1|1 typedef long long LL; int D,I; const int MAXN=1<<20; int tree[MAXN]; int ans; void query(int root,int d){ if(d==D){ ans=root; return; } if(!tree[root]){ tree[root]=1; query(lson,d+1); } else{ tree[root]=0; query(rson,d+1); } } int main(){ int T; while(SI(T),T!=-1){ T_T{ mem(tree,0); SI(D);SI(I); while(I--){ query(1,1); } printf("%d\n",ans); } } return 0; }
Miguel Revilla 2000-08-14
相关文章推荐
- Android应用架构
- 《近匠》Worktile王涛:典型MEAN架构下的团队协作工具
- Android WiFi--系统架构
- SGU102 Coprimes
- linux系统下安装Tomcat
- awk读取XML文件并格式化数据
- Apache Kafka:下一代分布式消息系统
- opencv2.4.9学习_加载并显示一个图像
- Linux命令之sed的详解
- hadoop学习;hadoop伪分布搭建
- hadoop 轻松时刻 hdfs漫画
- 欢迎使用CSDN-markdown编辑器
- Mac下启动Apache
- Mac OS X中配置Apache
- CPU与内存互联的架构演变
- OpenGL基于帧缓存FBO的离屏渲染
- 【Modern OpenGL】转换 Transformations
- 马哥Linux笔记--VMware Workstation系统虚拟化
- CentOS-7 安装 Tomcat8.X
- google 指定网站,关键字大小写结果竟然完全不同