ctype read finish car ann bbb over lines 遞歸

The End of The World

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 646 Accepted Submission(s): 308


Problem Description Legend says that there is a group of monks who are 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 are stacked on one of the pegs, and ordered from largest (on the bottom) to smallest (on the top). The object is to move this stack of disks to another peg, subject to two rules: 1) you can only move one disk at a time, and 2) you cannot move a disk onto a peg if that peg already has a smaller disk on it.
The monks believe that when they finish, the world will end. Suppose you know how far they’ve gotten. Assuming that the monks are pursuing the most efficient solution, how much time does the world have left?
Input There will be several test cases in the input. Each test case will consist of a string of length 1 to 63, on a single line. This string will contain only (capital) 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 smallest disk, and so on, until the last character, which tells the position of the largest disk. The character will be A, B or C, indicating which peg the disk is currently on. You may assume that the monks’ overall goal is to move the disks from peg A to peg B, and that the input represents a legitimate position in the optimal solution. The input will end with a line with a single capital X.
Output For each test case, print a single number on its own line indicating the number of moves remaining until the given Towers of Hanoi problem is solved. Output no extra spaces, and do not separate answers with blank lines. All possible inputs yield answers which will fit in a signed 64-bit integer.
Sample Input

AAA
BBB
X

Sample Output

7
0

Source The University of Chicago Invitational Programming Contest 2012
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Hanoi塔給一個中間狀態。求到終於狀態最小步數。

遞歸。

每次calc(n,goal) 表示把前n小的盤子全移動到goal的步數。

可遞歸分解為

{

1.若盤子n本來就在goal上，跳過

2.若盤子n不在goal上，

1)把前n-1個盤子移動到空余的goal2上 calc(n-1,goal2)

2)移動盤子n到goal 1步

3)把n-1個盤子從goal2移到goal上 2^(n-1)-1步

}

#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){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/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塔從中間狀態移動)