The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 217    Accepted Submission(s): 160

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)

题意：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。

分析：

1、这道题暴力打表其实可以找出规律

2、用dp来做就是设dp[i][0]表示在第i个格子里面不放木桩，dp[i][1]表示在第i个格子里面放木桩。

如果当前格子放木桩的话：那么dp[i][1]=dp[i-3][0]+dp[i-3][1]

如果当前格子不放木桩的话，那么dp[i][0]=dp[i-1][0]+dp[i-1][1]

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
const int MAXN=60+5;
LL dp[MAXN][2],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<=65;i++)
    {
        dp[i][0]=dp[i-1][0]+dp[i-1][1];
        dp[i][1]=dp[i-3][0]+dp[i-3][1]+1;
    }
}

int main()
{
    init();
    //freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF)
    {
        printf("%I64d\n",dp[n][0]+dp[n][1]);
    }
    return 0;
}