HDU 1041 Computer Transformation (简单大数)

Computer Transformation

http://acm.hdu.edu.cn/showproblem.php?pid=1041

Problem 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

 

 解题思路：找出规律，F[n] = F[n-2] + 2^n-3, 不过不能直接一起算，因为可能会超时；我们可以这样F[n] = F[n-2] + G[n-3]; (其中F[]代表多少对0，G[]代表有多少个1)；

G[n] = G[n-1] + G[n-1];所以这里就只需要大数加法就够了！

 

解题代码：

 #include <stdio.h>
 #include <iostream>
 #include <math.h>
 #include <string.h>
 #define CLE(name) memset(name, 0, sizeof (name))
 using namespace std;

 typedef __int64 LL;
 const int max_n = ;

 string F[*max_n];
 string P2[*max_n];
 int num1[max_n], num2[max_n];

 string add(string a, string b)
 {
     CLE(num1);
     CLE(num2);
     int len1 = a.length();
     int len2 = b.length();
     int k = ;
     for (int i = len1 - ; i >= ; i --)
     {
         num1[k++] = a[i] - '';
     }
     k = ;
     for (int i = len2 - ; i >= ; i --)
     {
         num2[k++] = b[i] - '';
     }
     int up = ;
     for (int i = ; i < max_n/; i ++)
     {
         num1[i] = num1[i] + num2[i] + up;
         up = num1[i]/;
         num1[i] %= ;

     }
     for (k = max_n - ; k >= ; k --)
     {
         if (num1[k] != )
             break;
     }
     a = "";
     for (; k >= ; k --)
         a += num1[k] + '';
     return a;
 }

 void pow2()
 {
     P2[] = "";
     P2[] = "";
     for (int i = ; i <= ; i ++)
     {
         P2[i] = add(P2[i-], P2[i-]);
     }
 }
 void deel()
 {
     F[] = "";
     F[] = "";
     F[] = "";
     for (int i = ; i <= max_n; i ++)
     {
         F[i] = add(F[i-], P2[i-]);
     }
     return;
 }

 int main()
 {
     int n;
     pow2();
     deel();
     while (~scanf ("%d", &n))
     {
         cout << F[n] << endl;
     }
     return ;
 }