UVA679 找规律

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

每个小球都会落在根结点上，因此前两个小球必然是一个在左子树，一个在右子树。一 般地，只需看小球编号的奇偶性，就能知道它是最终在哪棵子树中。对于那些落入根结点左 子树的小球来说，只需知道该小球是第几个落在根的左子树里的，就可以知道它下一步往左 还是往右了。依此类推，直到小球落到叶子上。 如果使用题目中给出的编号I，则当I是奇数时，它是往左走的第（I+1）/ 2个小球； 当I是偶数时，它是往右走的第I/ 2个小球。这样，可以直接模拟最后一个小球的路线。这样，程序的运算量就与小球编号无关了，而且节省了一个巨大的数组。

代码：

#include<bits/stdc++.h>const int INF  = 0x3f3f3f3f;const int Maxn = 100005;#define MST(s,q) memset(s,q,sizeof(s))#define Lchild id<<1#define Rchild (id<<1)+1using namespace std;int D, N, M, ans;void deal(int I, int time) {    if (time == D) return;    if (I % 2 == 1) {        ans *= 2;        deal((I + 1) / 2, time + 1);    } else {        ans = ans * 2 + 1;        deal(I / 2, time + 1);    }}int main() {    cin >> M;    while (M--) {        scanf("%d%d", &D, &N);        ans = 1;        deal(N, 1);        printf("%d\n", ans);    }    cin >> M;}#include<bits/stdc++.h>const int INF  = 0x3f3f3f3f;const int Maxn = 100005;#define MST(s,q) memset(s,q,sizeof(s))#define Lchild id<<1#define Rchild (id<<1)+1using namespace std;int D, N, M;int main() {    cin >> M;    while (M--) {        scanf("%d%d", &D, &N);        int k = 1;        for (int i = 0; i < D - 1; i++) {            if (N % 2) {                k = k * 2;                N = (N + 1) / 2;            } else {                k = k * 2 + 1;                N /= 2;            }        }        printf("%d\n", k);    }    cin >> M;}