HDU - 5366 The mook jong (dp动态规划)

The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 62    Accepted Submission(s): 40

Problem Description


ZJiaQ want to become a strong man, so he decided to play the mook jong。ZJiaQ want to put some mook jongs in his backyard. His backyard consist of n bricks that is 1*1,so it is 1*n。ZJiaQ want to put a mook jong in a brick. because of the hands of the mook jong， the distance of two mook jongs should be equal or more than 2 bricks. Now ZJiaQ want to know how many ways can ZJiaQ put mook jongs legally(at least one mook jong).

 

Input

There ar multiply cases. For each case, there is a single integer n( 1 < = n < = 60)

 

Output

Print the ways in a single line for each case.

 

Sample Input

   
   
   
   

    
    
    
    1	
2
3
4
5
6
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
3
5
8
12
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
 
     
The mook jong
  
     
 
       Accepts: 506 
     
  
     
 
       Submissions: 1281 
     
 
     
 
      
 
        Time Limit: 2000/1000 MS (Java/Others) 
      
  
      
 
        Memory Limit: 65536/65536 K (Java/Others) 
      
 
     
 
    
    
    
    
    
 
     

       问题描述 
     
 
     
ZJiaQ为了强身健体，决定通过木人桩练习武术。ZJiaQ希望把木人桩摆在自家的那个由1*1的地砖铺成的1*n的院子里。由于ZJiaQ是个强迫症，所以他要把一个木人桩正好摆在一个地砖上，由于木人桩手比较长，所以两个木人桩之间地砖必须大于等于两个，现在ZJiaQ想知道在至少摆放一个木人桩的情况下，有多少种摆法。
 
     

       输入描述 
     
 
     
输入有多组数据，每组数据第一行为一个整数n(1 < = n < = 60)
 
     

       输出描述 
     
 
     
对于每组数据输出一行表示摆放方案数
 
     

       输入样例 
     
 
     
1	
2
3
4
5
6
 
     

       输出样例 
     
 
     
1
2
3
5
8
12
 
    
    
    
    
    
/*
Author: 2486
Memory: 1404 KB		Time: 0 MS
Language: C++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
const int MAXN = 60 + 5;
LL dp[MAXN][2];
int n;
void init() {
    dp[1][0] = 0,dp[1][1] = 1;
    dp[2][0] = 1,dp[2][1] = 1;
    dp[3][0] = 2,dp[3][1] = 1;
    for(int i = 4; i < MAXN; i ++) {
        dp[i][0] = dp[i - 1][0] + dp[i - 1][1];//如果我此时这个位置不放桩子,那么状态可以从dp[i - 1][0] 以及dp[i - 1][1]传递
        dp[i][1] = dp[i - 3][0] + dp[i - 3][1] + 1;//如果放桩子,可以从前两个桩子前传递过来状态
    }
}
int main() {
    init();
    //freopen("D://imput.txt", "r", stdin);
    while(~ scanf("%d", &n)) {
        printf("%I64d\n", dp[n][0] + dp[n][1]);
    }
    return 0;
}