B

题意：

给出汉诺塔原型

求在第几步的时候这个汉诺塔每一根柱子上面的盘子数一样多

题解：

打表找规律

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int k;int s[10];int hanoi(char a,char b,char c,int n){    if(n==0)        return 0;    if(hanoi(a,c,b,n-1))        return 1;    --s[a-'a'],++s[b-'a'],s[c-'a'];    if(s[a-'a']==s[b-'a']&&s[b-'a']==s[c-'a']){        printf("%d\n",k);        return 1;    }    k++;    if(hanoi(c,b,a,n-1))        return 1;}int main(){    int n;    //freopen("out.txt","w",stdout);    for(int i=3;i<=30;i+=3)    {        k=1;        s[0]=i;        s[1]=s[2]=0;        hanoi('a','b','c',i);    }    return 0;}

得到规律：

num【3】= 2，num【6】=9，num【12】=38

当 i 为奇数的时候：num【i】=num【i-3】*4+2

当 i 为偶数的时候：num【i】=num【i-3】*4 - t【i】

其中 t【i】=t【i-6】*4+21

然后大整数处理一下就可以了

import java.io.File;import java.io.FileNotFoundException;import java.io.PrintWriter;import java.math.BigInteger;import java.util.Scanner;public class Main {public static void main(String[] args) throws FileNotFoundException {Scanner cin = new Scanner(new File("input.txt"));PrintWriter print = new PrintWriter(new File("output.txt"));int n;BigInteger []t = new BigInteger[307];BigInteger []num = new BigInteger[307];num[3] = BigInteger.valueOf(2);num[6] = BigInteger.valueOf(9);t[12] = BigInteger.valueOf(17);num[9] = BigInteger.valueOf(38);for(int i = 12;i<=300;i+=3){if((i%2)==0){num[i] = num[i-3].multiply(BigInteger.valueOf(4)).subtract(t[i]);t[i+6] = t[i].multiply(BigInteger.valueOf(4)).add(BigInteger.valueOf(21));}else {num[i] = num[i-3].multiply(BigInteger.valueOf(4)).add(BigInteger.valueOf(2));}}while(cin.hasNext()){n = cin.nextInt();print.println(num[n]);}cin.close();print.close();}}