Tower IX
Time Limit: 3000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 775 accepted submission (s): 462
Problem description1, 2,..., n indicates n plates. The number of plates is big. N are placed on 1st columns. The tray 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 lowest
Plate above. 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 the M-mobile plate.
Input two integers n (1 ≤ n ≤ 63) per line, m ≤ 2 ^ n-1.n = m = 0 exit
Output The number of plates that are moved MB.
Sample Input
63 163 20 0
Sample output
12
The idea is to first find the first few disks that can be completely moved at least, and then determine whether the disks are just completed. If the disks are just completed, the last one must be 1. If the disks are not completed, the next one is the next one completed this time, so we can determine if there is just one more. If there is one more, the last one will be it; otherwise, we will take the remaining number-1 (because we need to take a big one, if you don't understand it, think about it.) Repeat the above operation.
# Include <iostream>
# Include <cstring>
Using namespace STD;
Int N, lstn;
Long long M, ANS, num [10000];
Long long MYPOW (int x)
{
If (Num [x]) return num [x];
If (x = 0) return 1;
Num [x] = 2 * MYPOW (x-1 );
Return num [x];
}
Long long test (long mm ){
Lstn = 1;
While (MYPOW (lstn)-1 <mm ){
Lstn ++;
}
If (MYPOW (lstn)-1 = mm) return 0;
Else if (MYPOW (lstn-1) = mm) Return-1;
Else return mm-MYPOW (lstn-1 );
}
Int main (){
Memset (Num, 0, sizeof (Num ));
While (CIN> N> M & N ){
Ans = test (m );
While (ANS> 0 ){
Ans = test (ANS );
}
If (ANS = 0) cout <1 <Endl;
Else cout <lstn <Endl;
}
}
Hdu2175 tower IX
