The End of the worldTime
limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 646 Accepted Submission (s): 308
Problem Descriptionlegend says that there was a group of monks who was solving a large Towers of Hanoi puzzle. The Towers of Hanoi is a well-known puzzle, consisting of three pegs, with a stack of disks, each a different size. At the start, all of the disks is stacked on one of the pegs, and ordered from largest (on the bottom) to smallest (on th e top). The object is to move this stack of disks to another peg, subject-to-rules:1) you can only move one disk at a time, a nd 2) cannot move a disk onto a peg if that peg already have a smaller disk on it.
The monks believe that's when they finish and the world would end. Suppose know how far they ' ve gotten. Assuming that the monks is pursuing the most efficient solution, what much time does the world has left?
Inputthere'll is several test cases in the input. Each of the test case would consist of a string of length 1 to a. This string would contain only
AS
BS and
CS. The length of the string indicates the number of disks, and each character indicates the position of one disk. The first character tells the position of the smallest disk, the second character tells the position of the second Smalles t disk, and so on, until the last character, which tells the position of the largest disk. The character would be
A,
BOr
C, indicating which peg the disk is currently on. Assume that the monks ' overall goal are to move the disks from peg
Ato Peg
B, and that the input represents a legitimate position in the optimal solution. The input would end with a line with a single capital
X.
Outputfor each test case, print a single number on its own line indicating the number of moves remaining until the given T Owers of Hanoi problem is solved. Output no extra spaces, and do not separate answers with blank lines. All possible inputs yield answers which would fit in a signed 64-bit integer.
Sample Input
Aaabbbx
Sample Output
70
Sourcethe University of Chicago Invitational programming Contest 2012
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The Hanoi Tower gives an intermediate state. The minimum number of steps to finally state.
Recursion.
Each calc (n,goal) represents the number of steps to move the first n small plates to goal.
Can be recursively decomposed into
{
1. If the plate n is on the goal, skip
2. If the plate n is not on the goal,
1) Move the front n-1 plate to the Free Goal2 calc (n-1,goal2)
2) Move plate N to Goal 1 steps
3) Move n-1 plate from Goal2 to Goal 2^ (N-1)-1 steps
}
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include < functional> #include <iostream> #include <cmath> #include <cctype> #include <ctime>using namespace std; #define for (I,n) for (int. i=1;i<=n;i++) #define FORK (I,k,n) for (int. i=k;i<=n;i++) #define REP (I,n) for (int i=0;i<n;i++) #define ForD (I,n) for (int. i=n;i;i--) #define REPD (I,n) for (int. i=n;i>=0;i--) #define FORP (x) for ( int p=pre[x];p; p=next[p]) #define FORPITER (x) for (int &p=iter[x];p; p=next[p]) #define LSON (x<<1) #define Rson ((x<<1) +1) #define MEM (a) memset (A,0,sizeof (a)), #define MEMI (a) memset (A,127,sizeof (a)), #define MEMI (a) memset ( A,128,sizeof (a)); #define INF (2139062143) #define F (100000007) #define MAXN (63+10) typedef long Long Ll;ll Mul (ll A,ll b) {R Eturn (A*b)%F;} ll Add (ll A,ll b) {return (a+b)%F;} ll Sub (ll A,ll b) {return (A-b+llabs (a)/f*f+f)%F; void Upd (ll &a,ll b) {a= (a%f+b%f)%F;} Char s[maxn];int n;ll Calc (int n,int goal) {ForD (i,n{if (s[i]!=goal) {int goal2;if (S[i] + Goal = = ' a ' + ' B ') goal2= ' C '; if (S[i] + Goal = = ' a ' + ' C ') goal2= ' B '; if (S[i] + Goal = = ' C ' + ' B ') goal2= ' A '; return calc (i-1,goal2) + (1ll<<i-1);}} return 0;} int main () {//freopen ("hdu4260.in", "R", stdin), while (scanf ("%s", s+1) ==1) {if (s[1]== ' X ') break; N=strlen (s+1); cout <<calc (n, ' B ') <<endl;} return 0;}
HDU 4260 (The End of the World-hanoi tower moves from the middle state)