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 = 1002;

string F[2*max_n];
string P2[2*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 = 0;
    for (int i = len1 - 1; i >= 0; i --)
    {
        num1[k++] = a[i] - '0'; 
    }
    k = 0;
    for (int i = len2 - 1; i >= 0; i --)
    {
        num2[k++] = b[i] - '0'; 
    }
    int up = 0;
    for (int i = 0; i < max_n/2; i ++)
    {
        num1[i] = num1[i] + num2[i] + up;
        up = num1[i]/10;
        num1[i] %= 10;
    
    }
    for (k = max_n - 1; k >= 0; k --)
    {
        if (num1[k] != 0)
            break;
    }
    a = "";
    for (; k >= 0; k --)
        a += num1[k] + '0';
    return a;
}

void pow2()
{
    P2[0] = "1";
    P2[1] = "2";
    for (int i = 2; i <= 1000; i ++)
    {
        P2[i] = add(P2[i-1], P2[i-1]);
    }
}
void deel()
{
    F[0] = "0";
    F[1] = "0";
    F[2] = "1";
    for (int i = 3; i <= max_n; i ++)
    {
        F[i] = add(F[i-2], P2[i-3]);
    }
    return;
}

int main()
{
    int n;
    pow2();
    deel();
    while (~scanf ("%d", &n))
    {
        cout << F[n] << endl;
    }
    return 0;
}