UVA679模拟小球降落(大数模拟超时是找规律)

1)直接模拟，超时

#include <iostream>
#include <string.h>
using namespace std;
const int maxn=20;
int depth=0;int num=0;
int binary_tree[1<<maxn];//！左移符号，当左边的数为1时，其左移后相当于返回2^maxn
int found(int i)
{
int tem=1<<(depth-1);
if(i>=(tem)){
return i;
}
else
{
if(binary_tree[i]==-1){
binary_tree[i]*=-1;
found(i*2);
}
else{
binary_tree[i]*=-1;
found(i*2+1);
}
}
}
int main()
{
int kase;
while(cin>>kase&&kase!=-1){
for(int i=1;i<=kase;i++){
memset(binary_tree,-1,sizeof(binary_tree));
cin>>depth>>num;
for(int i=1;i<num;i++){
found(1);
}
cout<<found(1);
}
}
return 0;
}

2)模拟且超时，说明有规律：每一个节点处前后经过的两个球一定方向不一样，所以根据所求球的序号奇偶性模拟最后一个球的降落过程

#include <iostream>

using namespace std;
int main()
{
int n;
while(cin>>n&&n!=-1){
while(n--){
int dep,num;
cin>>dep>>num;
int k=1;
while(--dep){//while(dep--)先执行while里面的语句再--，反之先--
if(num%2){
<span style="white-space:pre">		</span>k=k*2;
num=num/2+1;
}
else{
k=k*2+1;
num=num/2;
}
}
cout<<k<<endl;
}
}
return 0;
}

3）用位移符号（第二个while里稍有变动）

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

int main()
{
int n,r,l;
while (cin >> n && n >= 0) {
while (n --) {
cin >> l >> r;
int k = 1;
while (-- l) {
if (r%2) k = k<<1;//k*=2
else k = (k<<1)+1;//位移符号优先级低，加括号
r = (r+1)>>1;//奇数/2向下取整，偶数/2
}
cout << k << endl;
}
}
return 0;
}

4）

Dropping Balls

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, eitherfalse ortrue. Initially, all of the flags
are false. When visiting a non-terminal node if the flag's current value at this node isfalse, then the ball will first switch this flag's value, i.e., from thefalse 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 thetrue to thefalse, 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 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 isD, the maximum depth of FBT, and the second one isI, the
Ith ball being dropped. You may assume the value ofI 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

Miguel Revilla

2000-08-14