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

2
3

Sample Output

1
1

Solution

題目的意思是找出有多少個連續的0，就是找規律看看。

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

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

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

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

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

算着的時候，我就發現0*2+1 = 1， 1*2-1 = 1， 1*2+1 = 3， 3*2-1 = 5，蠻符合的，就推論n=6的時候爲5*2+1 = 11，發現沒錯。

於是就得出了f(n) = f(n-1)*2 + (-1)^n，注意n有1000之大，所以要用高精度乘法，每次乘最多進一位，於是位數最多也就1000，敲出來就A啦

（PS：其實加一減一是看一下前一個序列首尾是否相同的時候，列表並綜合結果發現的）

#include <iostream>
#include <cstring>

using namespace std;

int ans[1005][1005];//高精度乘法的結果數組

int main()
{
  int i, j, n;

  memset(ans, 0, sizeof(ans));
  ans[1][0] = 0;
  for (int i = 2; i < 1005; ++i)//簡單粗暴的乘法，可以優化的
  {
    for (j = 0; j < 1005; ++j) ans[i][j] = ans[i-1][j] * 2;

    if (i%2) ans[i][0] -= 1;//處理加一減一
    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;//輸出0的時候的控制
    while (i >= 0) cout << ans[n][i--];
    cout << endl;
  }

  return 0;
}