标签：point terminal sum from cti ini ace max visit
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.
? 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 #
Line l + 1 Dl Il test case #l
Line l + 2 -1 a constant ‘-1’ representing the end of the input file
搬题太麻烦了，直接去uva看原题吧。
直接先上来来一遍模拟，3层循环，还有个2的20次方的数组，肯定过不了。
#include<cstdio>
#include<cstring>
bool s[1<<20] = {false};
using namespace std;
int main(){
int l;
scanf("%d",&l);
for(int i = 0; i < l; ++i){
int d,n,tempp = 1;
memset(s, false,sizeof(s));
scanf("%d%d",&d,&n);
for(int j = 0; j < n; j++){
tempp = 1;
for(int k = 1; k < d; k++){
int point = tempp;
if(s[tempp]==false){
tempp = 2 * tempp;
}else{
tempp = 2 * tempp + 1;
}
s[point] = !s[point];
if(j == n-1&&k==d-1){
printf("%d\n",tempp);
}
}
}
}
int a;
scanf("%d",&a);
return 0;
}
如我所料。
按紫书的写法，想办法让循环消掉一层。
#include<cstdio>
using namespace std;
int main(){
int l;
scanf("%d",&l);
for(int i = 0; i < l; ++i){
int d,n,k=1;
scanf("%d%d",&d,&n);
for(int j = 1; j < d; j++){
if(n%2){
k = k*2;
n = (n+1)/2;
}else{
k = k*2 + 1;
n /= 2;
}
}
printf("%d\n",k);
}
int a;
scanf("%d",&a);
return 0;
}
当这个球是通过该节点的奇数个数的球时，朝左走，偶数朝右走。又可以继续算出它是通过子节点的第几个。
这样就可以免去模拟前n-1个球掉落的全过程。
直接ac。
标签：point terminal sum from cti ini ace max visit
原文地址：https://www.cnblogs.com/ranbom/p/12961957.html
