1028. Hanoi Tower Sequence n足够大 第p步移动哪个木块

http://sicily.3322.org/show_problem.php?pid=1028 1028. Hanoi Tower Sequence
	Total: 4654 Accepted: 1115 Rating: 2.9/5.0(15 votes) 012345

	Time Limit: 1s    Memory Limit: 32MB Description Hanoi Tower is a famous game invented by the French mathematician Edourard Lucas in 1883. We are given a tower of n disks, initially stacked in decreasing size on one of three pegs. The objective is to transfer the entire tower to one of the other pegs, moving only one disk at a time and never moving a larger one onto a smaller.  The best way to tackle this problem is well known: We first transfer the n-1 smallest to a different peg (by recursion), then move the largest, and finally transfer the n-1 smallest back onto the largest. For example, Fig 1 shows the steps of moving 3 disks from peg 1 to peg 3. Now we can get a sequence which consists of the red numbers of Fig 1: 1, 2, 1, 3, 1, 2, 1. The ith element of the sequence means the label of the disk that is moved in the ith step. When n = 4, we get a longer sequence: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. Obviously, the larger n is, the longer this sequence will be. Given an integer p, your task is to find out the pth element of this sequence. Input The first line of the input file is T, the number of test cases. Each test case contains one integer p (1<=p<10^100). Output Output the pth element of the sequence in a single line. See the sample for the output format. Print a blank line between the test cases. Sample Input Copy to clipboard 4 1 4 100 100000000000000 Sample Output Case 1: 1 Case 2: 3 Case 3: 3 Case 4: 15

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=600;
int t[maxn],A[maxn];
//{ 0-9,A-Z,a-z } 62中符号
void change(int a,int b,char *s,char *d)//s在a进制下，转换成b进制，保存到d中
{
    int i,k,l;
    for(k=i=strlen(s);0<i--;)
                     t[k-1-i]=s[i]-(s[i]<58?48:s[i]<97?55:61);
            for(l=0;k;)
            {
                     for(i=k;1<i--;)
                     {
                               t[i-1]+=t[i]%b*a;
                               t[i]/=b;
                     }
                     A[l++]=t[0]%b;
                     t[0]/=b;
                     for(;0<k&&!t[k-1];k--);
            }
            for(d[l]=i=0;i<l;i++)
                     d[l-1-i]=A[i]+(A[i]<10?48:A[i]<36?55:61);
}
int main()
{
    int ci;scanf("%d",&ci);
    for(int pl=1;pl<=ci;pl++)
    {
        if(pl>1) printf("/n");
        char a[maxn],b[maxn];
        scanf("%s",a);
        change(10,2,a,b);
        int cnt=1;
        for(int i=strlen(b)-1;i>=0&&b[i]=='0';i--) cnt++;
        printf("Case %d: %d/n",pl,cnt);
    }
    return 0;
}