Sicily 1166 Computer Transformat

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simultaneously tranforms each digit 0 into the sequence 1 0 and each digit 1 into the sequence 0 1. So, after the first time step, the sequence 0 1 is obtained; after the second, the sequence 1 0 0 1, after the third, the sequence 0 1 1 0 1 0 0 1 and so on.

How many pairs of consequitive zeroes will appear in the sequence after n steps?

Input

Every input line contains one natural number n (0 < n ≤1000).

Output

For each input n print the number of consecutive zeroes pairs that will appear in the sequence after n steps.

Sample Input

23

Sample Output

11

Solution

题目的意思是找出有多少个连续的0，就是找规律看看。

n=1  ---------> 01                                                                    --------> 0

n=2  ---------> 1001                                                                --------> 1

n=3  ---------> 01101001                                                       --------> 1

n=4  ---------> 1001011001101001                                     --------> 3

n=5  ---------> 01101001100101101001011001101001 --------> 5

算着的时候，我就发现0*2+1 = 1， 1*2-1 = 1， 1*2+1 = 3， 3*2-1 = 5，蛮符合的，就推论n=6的时候为5*2+1 = 11，发现没错。

于是就得出了f(n) = f(n-1)*2 + (-1)^n，注意n有1000之大，所以要用高精度乘法，每次乘最多进一位，于是位数最多也就1000，敲出来就A啦

（PS：其实加一减一是看一下前一个序列首尾是否相同的时候，列表并综合结果发现的）

#include <iostream>#include <cstring>using namespace std;int ans[1005][1005];//高精度乘法的结果数组int main(){  int i, j, n;  memset(ans, 0, sizeof(ans));  ans[1][0] = 0;  for (int i = 2; i < 1005; ++i)//简单粗暴的乘法，可以优化的  {    for (j = 0; j < 1005; ++j) ans[i][j] = ans[i-1][j] * 2;    if (i%2) ans[i][0] -= 1;//处理加一减一    else ans[i][0] += 1;    for (j = 0; j < 1005; ++j) if (ans[i][j] > 9)    {      ans[i][j+1] += ans[i][j] / 10;      ans[i][j] %= 10;    }  }  while (cin >> n)  {    i = 1004;    while (ans[n][i] == 0 && i > 0) --i;//输出0的时候的控制    while (i >= 0) cout << ans[n][i--];    cout << endl;  }  return 0;}