The mook jong

The mook jong

Accepts: 506

Submissions: 1281

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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
题意：问题可以转化成在一个数轴上，隔不低于两个点放至少一个木桩的方法。那么放或是不放点就在那里，你有几种方法放置木桩？（木桩数量不限

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <cstring>

using namespace std;

#define INF 0xfffffff
#define maxn 10005

int main()
{
    __int64 n, dp[maxn] = {1, 2, 3, 5};

    while(scanf("%I64d", &n) != EOF)
    {
        for(int i = 3; i <= n; i++)
            dp[i] = dp[i-3] + dp[i-1];
        printf("%I64d
", dp[n]-1);  // 减去一种什么都不放的情况

    }
    return 0;
}