HDOJ2175 汉诺塔IX[递推]

汉诺塔IX

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 437    Accepted Submission(s): 250

Problem Description

1,2,...,n表示n个盘子．数字大盘子就大．n个盘子放在第１根柱子上．大盘不能放在小盘上．
在第１根柱子上的盘子是a[1],a[2],...,a[n]. a[1]=n,a[2]=n-1,...,a[n]=1.即a[1]是最下
面的盘子．把n个盘子移动到第３根柱子．每次只能移动１个盘子，且大盘不能放在小盘上．
问第m次移动的是那一个盘子.

 

Input

每行2个整数n (1 ≤ n ≤ 63) ,m≤ 2^n-1.n=m=0退出

 

Output

输出第m次移动的盘子的号数．

 

Sample Input

63 1 63 2 0 0

 

Sample Output

1 2

 

Author

zhousc

 

Source

ECJTU 2008 Summer Contest

 

Recommend

lcy

 

 

 

 

规律：

 

121

1213121

121312141213121

1213121412131215121312141213121

 

 

2^(n-2)+(k-1)*(2^(n-1))=(2*k-1)*(2^(n-1))

2*k-1为奇数

code:

#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    __int64 n,m;
    int i;
    while(~scanf("%I64d%I64d",&n,&m),n||m)
    {
        if(m%2)
        {
            printf("1\n");
            continue;
        }
        if(m==int(1<<(n-1)))
        {
            printf("%I64d\n",n);
            continue;
        }
        i=1;
        while(m%2==0)
        {
            m/=2;
            i++;
        }
        printf("%d\n",i);
    }
    return 0;
}