Nudnik Photographer URAL - 1260 [规律DP]

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:

The 1 year old student is to sit at the left end of the row.
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

input	output
4	4

Notes

If N = 4 then there are following possible arrangements: (1,2,3,4), (1,2,4,3), (1,3,2,4) and (1,3,4,2).

Problem Author: Alexander Ipatov
Problem Source: Open collegiate programming contest for high school children of the Sverdlovsk region, October 11, 2003

Tags: dynamic programming ( hide tags for unsolved problems)

Difficulty: 118 Printable version Submit solution Discussion (9)
All submissions (9147) All accepted submissions (4352) Solutions rating (3672)

/*
给1-n<=55个数,对这些数进行排列，
要求：
    1必须放在最左边，
    相邻两个数必须差绝对值小于2；
    那么
    当1,2则有dp[i - 1]种组合;
    当1,3,2则是dp[i - 3]推过来;
    最后加上1357..642
    dp[i] = dp[n - 1] + dp[n - 3] + 1;
*/

#include<bits/stdc++.h>
#define ll long long

using namespace std;
const int maxn = 100;
ll dp[maxn];
int n;

void init()
{
    memset(dp, 0, sizeof(dp));
    dp[1] = 1; dp[2] = 1; dp[3] = 2; dp[4] = 4;
    for(int i = 5; i <= 55; i++) dp[i] += dp[i - 3] + dp[i - 1] + 1;
}

int main()
{
    init();
    while(~scanf("%d",&n))
    {
        printf("%lld\n", dp[n]);
    }
    return 0;
}