杭电OJ——1041 Computer Transformation

Computer Transformation

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
   
   
   
   

 

Source

Southeastern Europe 2005

 

Recommend

JGShining

 

1 题目意思很明确：step 0：1

step 1：01
step 2：1001
step 3：0110 1001
step 4：1001 0110 0110 1001
step 5：0110 1001 1001 0110 1001 0110 0110 1001...通过自己推算，我们可以发现规律：从4开始，第n串中00的来源主要有两部分：
1. 第n-2串中00的数目
2. 第n-2串中1的数目而第n串中1的数目=2^(n-1)
因此，递推公式为：

C(n)=C(n-2)+2^(n-3) 其中n>=4。

 其中用到了大数的处理，整体来说不难！我们先打表，之后再输出即可！话说这杭电第一页上的题目怎么有那么多水题，受不了了，下一次做一点有技巧性的题目吧！

代码如下：

#include <iostream>
#include <string>
using namespace std;
const int SIZE = 1001;
const int BASE = 10;
string result[SIZE];
string two("2");

void initial()
{
	int mcarry, temp, carry;
	int k;
	string tempResult;
	result[1] = "0";
	result[2] = "1";
	result[3] = "1";
	result[4] = "3";
	for (int i = 5; i < SIZE; ++i)
	{
		mcarry = 0;
		for (int j = 0; j < two.length(); ++j)  /*先做乘法,求出2^(n-3)*/
		{
			temp = 2 * (two[j] - '0') + mcarry;
			mcarry = temp / BASE;  /*乘法进位*/
			temp = temp % BASE;    
			tempResult += (temp + '0');
		}
		if (mcarry != 0)  /*进位不为0*/
		{
			tempResult += (mcarry + '0');
		}

		two = tempResult;  /*存储计算结果*/
		tempResult.clear();

		int minLength = two.length() > result[i - 2].length() ?  result[i - 2].length() : two.length();
		carry = 0;

		for (k = 0; k < minLength; ++k)   /*然后做加法f(n) = f(n -2) + 2^(n - 3)*/
		{
			temp = (two[k] - '0') + (result[i - 2][k] - '0') + carry;
			carry = temp / BASE;  /*加法进位*/
			temp = temp % BASE;
			result[i] += (temp + '0');
		}
		/*两个数可能长短不一,所以要比较再相加*/
		if (minLength < two.length())
		{
			for(; k < two.length(); k++)
			{
				temp = (two[k] - '0') + carry;
				carry = temp / BASE;
			    temp = temp % BASE;
			    result[i] += (temp + '0');
			}
		}
		if(minLength < result[i - 2].length())
		{
            for(; k < result[i - 2].length(); ++k)
			{
				temp = (result[i - 2][k] - '0') + carry;
				carry = temp / BASE;
			    temp = temp % BASE;
			    result[i] += (temp + '0');
			}
		}
		if (carry != 0)   /*进位不为0*/
		{
			result[i] += (carry + '0');
		}
		tempResult.clear();
	}
}
int main()
{
	int n;
	initial();  /*先打表*/
	while (scanf ("%d", &n) != EOF)
	{
		for(int i = result[n].length(); i > 0; --i)
		cout << result[n][i - 1];   /*这一步很关键，由于数据是倒着存的,所以要反着输出*/
		cout << endl;
	}
	system ("pause");
	return 0;
}