JOJ-2033-Towers of Hanoi

Tomorrow is Kate's birthday, Kate is a lovely girl aged 20. This Contest is held for her birthday and I want everyone will be happy today. My name is sea .I'm not so good at English and Programming. But when I know that it is possible to have a personal contest on JOJ. I want to have a try. Just for her.

Kate : Wish you beauty and healthy forever

happy birthday!

Do you know the game "Towers of Hanoi"? It is a very famous game. But people stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task ! Kate loves the game. She want to know how many times she has to move the disks at least to complete the game?

As we all know , for n disks the game takes 2^n-1 moves at least. But kate want to know the exact numble. Sea want to tell her ,but he is poor at math. So he want to write a program to help her.

Input Specification

Each line contains a numbe 'N' (0 <= N <= 500). N stand for the number of the disks.

Output Specification

Just output the least time that kate has to move the disks to complete the game. one per line.

Sample Input

7
10

Sample Output

127
1023

#include<iostream>
using namespace std;

const int MAX=501;
const int D=1000000000;
const int width=9;
void main()
{
int fact[MAX][17]={0};//151/9==16.777
int len[MAX]={0};
fact[0][0]=1;
len[0]=1;
for(int i=1;i<MAX;++i)
{
int c=0;
int idx=0;
for(idx=0;idx<len[i-1];++idx)
{
int t=fact[i-1][idx]*2+c;
fact[i][idx]=t%D;
c=t/D;
}
if(c>0)
{
len[i]=len[i-1]+1;
fact[i][idx]=c;
}
else
len[i]=len[i-1];
}
int n;
while(cin>>n)
{
if(len
==1)
{
cout<<fact
[len
-1]-1<<endl;
continue;
}
cout<<fact
[len
-1];
for(int i=len
-2;i>0;--i)
{
cout.width(width);
cout.fill('0');
cout<<fact
[i];
}
cout.width(width);
cout.fill('0');
cout<<fact
[0]-1;
cout<<endl;
}
}

　
　

说简单点就是用数组计算2的乘幂　两点教训：第一，要计算2^0。在这道题里可能很自然考虑到，然而在一道计算N!的程
序可难住了我，郁闷了好久。实际上这个程序就是用计算N!的程序改。第二，不要以为加法比乘法快。开始我是用加法，
用时0.01秒，而用乘法是0.00秒，做移位运算毕竟要比加法快多了。