1041

Posted 2022-11-28 tookkke

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7258    Accepted Submission(s): 2652


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

    找规律+大数运算（这里就直接使用自己写好的模版了，才不是偷懒呢）

    递推公式为F[n]=F[n-1]+2*F[n-2];

    可以求出通项公式是 F[n]=(2^(n-1)+(-1)^n)/3;

    关于模版的在http://blog.csdn.net/tookkke/article/details/51366490

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include "longint.h"

using namespace std;

const longint a=2,b=3,c=1;
int n;

int main()

	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	while(scanf("%d",&n)==1)
	
		if(n&1)cout<<((a^(n-1))-c)/b;
		else cout<<((a^(n-1))+c)/b;
		putchar('\\n');
	
	return 0;