soj1166. Computer Transformat（dp + 大数相加）

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

本来还以为是一道基本的动态规划题目，结果做了一遍WA，然后仔细观察，到了1000时数太大了，long long都放不下，所以尝试大数相加，一次就过了。。

思路：

dp[i][0] —— 第i轮后出现多少个01

dp[i][1] —— 第i轮后出现多少个1

dp[0][0] = 0; dp[0][1] = 1;

dp[1][0] = dp[1][1] = 1;

dp[i][0] = dp[i-2][0] + dp[i-1][1];

dp[i][1] = 2*dp[i-1][1];

n = dp[n-1][0];

#include <iostream>
#include <string>
using namespace std;

string dp[1001][2];

string add(string a,string b)
{
	string result;
	string rr;
	int i;
	int l1,l2,len1,len2;
	l1 = len1 = a.size();
	l2 = len2 = b.size();
	int aa,bb,cc,dd;
	dd = 0;
	while(l1 > 0 && l2 > 0)
	{
		aa = a[l1-1] - '0';
		bb = b[l2-1] - '0';
		cc = (aa + bb+dd) % 10;
		dd = (aa + bb+dd) / 10;
		result += cc+'0';
		l1--;
		l2--;
	}
	while(l1 > 0)
	{
		aa = a[l1-1] - '0';
		cc = (aa + dd) % 10;
		dd = (aa + dd) / 10;
		result += cc + '0';
		l1--;
	}
	while(l2 > 0)
	{
		bb = b[l2-1] - '0';
		cc = (bb + dd) % 10;
		dd = (bb + dd) / 10;
		result += cc + '0';
		l2--;
	}
	if(dd == 1)
		result += '1';
	for(i = result.size() - 1;i >= 0 ;i--)
		rr += result[i];
	return rr;
}

void init()
{
	int i;
	dp[0][0] = "0";
	dp[0][1] = "1";
	dp[1][0] = dp[1][1] = "1";
	for(i = 2;i <= 1000;i++)
	{
		dp[i][0] = add(dp[i-2][0],dp[i-1][1]);
		dp[i][1] = add(dp[i-1][1],dp[i-1][1]);
	}
}

int main()
{
	int n;
	init();
	while(cin >> n)
	{
		cout << dp[n-1][0] << endl;
	}
	return 0;
}