UVA 679

Dropping Balls

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

Submit Status

Description

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: 

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 

54 23 410 12 28 128-1

Sample Output 

1275123255

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Miguel Revilla
2000-08-14

        4 2 的示例的图

先说一下题意吧，给你一棵深度为D的满二叉树，给你I个球，初始所有的节点都是关闭的，如果节点是关闭的则球向左走，否则向右走，每个节点被球接触后节点开关被打开。我开始是直接模拟的，超时，以下为超时代码

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int tree[1 << 21];int main(){#ifdef CDZSCfreopen("i.txt", "r",stdin);#endifint n,a,b;while (~scanf("%d", &n)){if (n == -1)break;while (n--){scanf("%d%d", &a, &b);memset(tree, 0, sizeof(tree));int ans;while (b--){int temp = 1;for (int i = 0;i<a;i++){ans = temp;if (tree[temp]){temp = (temp << 1|1);}else{tree[temp] = 1;temp = temp << 1;}}}printf("%d\n", ans);}}return 0;}

这里的复杂度是O(D*I*N)，N最大是1W，最坏会达到接近10^9，这时就需要找规律了.我们可以想一下一个球掉下如果是第一次掉下则向左走，第二次向右走那么根据小球编号的奇偶就可以判定它所在的子树：I为奇数它是向左走的第（i+1）/2；偶数则是右走的（I）/2个小球

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){#ifdef CDZSCfreopen("i.txt", "r",stdin);#endifint n,a,b;while (~scanf("%d", &n)){if (n <= -1)break;while (n--){scanf("%d%d", &a, &b);int ans = 1;for (int i = 0; i < a - 1; i++){if (b % 2){ans <<= 1;//向左走b = (b + 1) >> 1; }else { ans = ans << 1 | 1; //向右走b >>= 1; }}printf("%d\n", ans);}}return 0;}