Nudnik Photographer

1260. Nudnik Photographer

Time limit: 1.0 second
Memory limit: 64 MB

If two people were born one after another with one second difference and one of them is a child, then the other one is a child too. We get by induction that all the people are children.

Everyone knows that the mathematical department of the Ural State University is a big family of Npersons, 1, 2, 3, …, N years old respectively.

Once the dean of the department ordered a photo if his big family. There were to be present all the students of the department arranged in one row. At first the dean wanted to arrange them by their age starting from the youngest student, but than he decided that it would look unnatural. Than he advised to arrange the students as follows:

1. The 1 year old student is to sit at the left end of the row.
2. The difference in ages of every two neighbors mustn’t exceed 2 years.

The dean decided that thereby the students would seem look as they were arranged by their ages (one can hardly see the difference in ages of 25 and 27 years old people). There exist several arrangements satisfying to the requirements. Photographer didn’t want to thwart dean’s desire and made the photos of all the possible mathematical department students’ arrangements.

Input

There is the integer number N, 1 ≤ N ≤ 55.

Output

the number of photos made by the photographer.

Sample

inputoutput

4

4

#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <vector>#include <bitset>#include <cstdio>#include <string>#include <numeric>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#include <functional>using namespace std;typedef long long  ll;typedef unsigned long long ull;int dx[4]={-1,1,0,0};int dy[4]={0,0,-1,1};//up down left rightbool inmap(int x,int y,int n,int m){if(x<1||x>n||y<1||y>m)return false;return true;}int hashmap(int x,int y,int m){return (x-1)*m+y;}#define eps 1e-8#define inf 0x7fffffff#define debug puts("BUG")#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define read freopen("in.txt","r",stdin)#define write freopen("out.txt","w",stdout)#define maxn 60long long dp[maxn];//dp[j]=dp[j-1]+1+dp[j-3]/**   第一个位置只能放1，则第二个位置可以放2或者3，因此，   1，当第二个位置放2时，有1 （2……），括号里的为2~n，就相当于1~n-1，即dp[n-1]   2,当第二个位置放3时，有1 3 （4……）2和1 3 2 （4……）两种，     2.1，第一种只可能是 1 3 4 6 …… 7 5 2这一种，     2.2，1 3 2 （4……）括号里就相当于1~n-3，即dp[n-3]所以，dp[n]=dp[n-1]+1+dp[n-3]**/int main(){    int n;    scanf("%d",&n);    dp[0]=0;dp[1]=1;dp[2]=1;    for(int i=3;i<=n;i++)        dp[i]=dp[i-1]+1+dp[i-3];    printf("%lld\n",dp[n]);    return 0;}