hdu 5366 The mook jong 动态规划

The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 402    Accepted Submission(s): 306

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
   
   
   
   

 

Source

BestCoder Round #50 (div.2)

枚举最后放的位置，最后放的位置要求前两个位置没有放，所以累加0-i-3即可

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
ll dp[80];
ll ans[80];
void work(int n){
    if(ans[n] != -1) return ;
    memset(dp,0,sizeof(dp));
    dp[0] = 1,dp[1] = 1,dp[2] = 1;
    for(int i = 3;i <= n;i++){
        for(int j = i-3;j>=0;j--)
            dp[i] += dp[j];
    }
    ans[n] = 0;
    for(int i = 1;i <= n; i++)
        ans[n] += dp[i];
}
int main(){
    int n;
    memset(ans,-1,sizeof(ans));
    while(scanf("%d",&n)!=EOF){
        work(n);
        cout<<ans[n]<<endl;
    }
    return 0;
}