HDU2511: tower X

Problem Description
1, 2 ,..., n indicates n plates. the number is big. place n plates on 1st columns. the dashboard cannot be placed on a small disk. the plates on the 1st columns are a [1], a [2],..., a [n]. a [1] = n, a [2] = n-1 ,..., a [n] = 1. that is, a [1] is the bottom plate. move n plates to 3rd columns. only one plate can be moved at a time, and the tray cannot be placed on a small disk. ask which plate is moved at MB and which column is moved from. for example, n = 3, m = 2. the answer is: 2 1 2, that is, the disk is moved from 1st to 2nd.

 

Input
The fifth row is an integer T, indicating that there are T groups of data. There are T rows below. Each row has two integers n (1 ≤ n ≤ 63), m ≤ 2 ^ n-1

 

Output
Output The number of plates and the number of columns that move the MB number.

 

Sample Input
4
3 2
4 5
39 183251937942
63 3074457345618258570

Sample Output
2 1 2
1 3 1
2 2 3
2 2 3

First, list the situation and find the rule.

Rule is

If the total number of disks is the same as that of the current number, the order of movement is 1-> 3-> 2-> 1.

Otherwise it is 1-> 2-> 3-> 1

Remember that the moving order is for the same disk.

 

[Cpp] view plaincopyprint? # Include <stdio. h>
 
_ Int64 pow (_ int64 n)
{
_ Int64 s = 1;
Int I;
For (I = 1; I <= n; I ++)
S * = 2;
Return s;
}
 
Int main ()
{
_ Int64 n, m, t;
Int a [3] [2] = {3, 1}, {1, 2}, {2, 3}, B [3] [2] = {2, 1}, {1, 3 }, {3, 2 }};
Int I, test;
Scanf ("% d", & test );
While (test --)
{
_ Int64 cnt = 0;
Scanf ("% I64d % I64d", & n, & m );
_ Int64 tem = m;
For (I = 1; I <= n; I ++)
{
T = m % 2;
If (t)
{
Printf ("% d", I );
Break;
}
Else
M/= 2;
}
If (tem % 2)
Cnt = (tem + 1)/2;
Else
{
M = tem;
_ Int64 sum = pow (I-1 );
M = m-sum;
Cnt = m/pow (I) + 1;
}
Int flag = 0;
If (n % 2 & I % 2) | (n % 2 = 0 & I % 2 = 0 ))
{
Flag = 1;
}
Cnt = cnt % 3;
If (flag)
{
Printf ("% d \ n", B [cnt] [0], B [cnt] [1]);
}
Else
{
Printf ("% d \ n", a [cnt] [0], a [cnt] [1]);
}
}
 
Return 0;
}

# Include <stdio. h>

_ Int64 pow (_ int64 n)
{
_ Int64 s = 1;
Int I;
For (I = 1; I <= n; I ++)
S * = 2;
Return s;
}

Int main ()
{
_ Int64 n, m, t;
Int a [3] [2] = {3, 1}, {1, 2}, {2, 3}, B [3] [2] = {2, 1}, {1, 3 }, {3, 2 }};
Int I, test;
Scanf ("% d", & test );
While (test --)
{
_ Int64 cnt = 0;
Scanf ("% I64d % I64d", & n, & m );
_ Int64 tem = m;
For (I = 1; I <= n; I ++)
{
T = m % 2;
If (t)
{
Printf ("% d", I );
Break;
}
Else
M/= 2;
}
If (tem % 2)
Cnt = (tem + 1)/2;
Else
{
M = tem;
_ Int64 sum = pow (I-1 );
M = m-sum;
Cnt = m/pow (I) + 1;
}
Int flag = 0;
If (n % 2 & I % 2) | (n % 2 = 0 & I % 2 = 0 ))
{
Flag = 1;
}
Cnt = cnt % 3;
If (flag)
{
Printf ("% d \ n", B [cnt] [0], B [cnt] [1]);
}
Else
{
Printf ("% d \ n", a [cnt] [0], a [cnt] [1]);
}
}

Return 0;
}