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] = {, , , };

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

     }
     return ;
 }