Topic 1113: Binary Tree

Time limit:1 second

Memory limit:32 MB

Special Judgment:No

Submit:3332

Solution:983

Description:

 

 

 

As shown above, it is composed of positive integers 1, 2, 3 ...... A special binary tree. We know that the last node of this binary tree is N. The problem is how many nodes are in the subtree where node m is located.

For example, if n = 12, M = 3, the nodes 13, 14, 15 and the following do not exist. the nodes in the subtree where node m is located include 3, 6, 7, 12, therefore, the subtree where node m is located has four nodes.

Input:

The input data includes multiple rows. Each row provides a set of test data, including two integers, M and N (1 <= m <= n <= 1000000000 ). The last group of test data contains two zeros, indicating the end of the input, which does not need to be processed.

Output:

For each group of test data, output a row. The row contains an integer that shows the number of nodes in the Child tree where node m is located.

Sample input:

3 120 0

Sample output:

4

Source:

Question about the vitality of computer research at Peking University in 2007

The problem is not difficult, but the recursion times out. You can calculate the number of layers and use the formula to calculate the number of points.

There is a recursion problem, that is, after each group of data is complete, because time is a global variable, so it will not be set to zero next time. If it is cleared in recursion, it will also kneel down. Therefore, after calling Main (), time = 0. This is acceptable because of the time global variable.

Recursive code:

 1 #include <stdio.h> 2 int time=0; 3 int timecount(int m,int n) 4 {     5     if(m<=n) 6         time++;     7     else 8         return 0; 9     timecount(2*m,n);    10     timecount(2*m+1,n);11     12     return time;13 }14 int main(void)15 {16     int m,n,t;17     18     while(scanf("%d %d",&m,&n)!=EOF&&m!=0)19     {    20         t=timecount(m,n);21         time=0;22         printf("%d\n",t);23     }24     return 0;25 }

 

Topic 1113: Binary Tree