The mook jong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 226    Accepted Submission(s): 167

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

题目大意：在由1*1的地砖铺成的1*n的院子里摆放木桩，每个木桩放在一个地板上。每两个木桩之间间隔至少两个地板，问有多少种放法。

解题思路：比赛的时候用的搜索，那叫一个慢，就打了40——60之间的表，算是蒙混过关。但是其实正解是dp。对于某个地板，可以放或者不放，那么如果放的话，结果加上dp[i-3]。如果不放的话，dp[i-1]是最优子结构，同时加上一共只放一个木桩在i位置的情况。dp[i]=dp[i-3]+dp[i-1]+1。结果超int需注意。

#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
INT dp[65];
int main(){
    dp[1]=1,dp[2]=2,dp[3]=3;
    for(int i=4;i<=62;i++)
        dp[i]=dp[i-3]+dp[i-1]+1;
    int n;
    while(scanf("%d",&n)!=EOF){
        printf("%lld\n",dp[n]);
    }
    return 0;
}