hihocoder-1732-1 - Bias permutation

hihocoder-1732-1 - Bias permutation 

#1732 : 1-bias permutation

Time limit: 10000ms

Single point time limit: 1000ms

Memory limit: 256MB

describe

If any element Pi in a 1~N permutation P=[P1, P2, ... PN] satisfies |Pi-i| ≤ 1, we call P a 1-bias permutation.

Given an N, please calculate how many different permutations are 1-bias permutations.

For example for N=3, there are 3 1-bias permutations: [1, 2, 3], [1, 3, 2], [2, 1, 3].

enter

an integer N.  

For 50% of the data, 1 ≤ N ≤ 15  

For 100% of the data, 1 ≤ N ≤ 50

output

an integer representing the answer

sample input

3

Sample output

3

 

 

Find the law, a number i can only exist in the three positions i-1, i, i+1.

dp[n][0] indicates the number of legal array types with n at position n, dp[n][1] indicates the number of legal array types with n at position n-1 and n at position n-1

　　For an array of original n elements, to insert an n+1, there are two insertion methods,

Put it at position n+1, then dp[n+1][0] = dp[n][0] + dp[n][1]; 

Put it at position n-1, then dp[n+1][1] = dp[n][0]; 

 

The sequence of such numbers is the Fibonacci sequence. f[n] = f[n-1] + f[n-2]; a simplified version of this.

 

#include <cstdio>
#include <cstdlib>
const int MAXN = 100 + 10;

int n;
long long ans, dp[MAXN][2];

void init(){
	dp[1][0] = 1;
	dp[1][1] = 0;  
	for(int i=2; i<MAXN; ++i){
		dp[i][0] = dp[i-1][0] + dp[i-1][1];
		dp[i][1] = dp[i-1][0];
	}
}

int main(){

	init();  
	while(scanf("%d", &n) != EOF){
		ans = dp[n][0] + dp[n][1];

		printf("%lld\n", ans );
	}  
	return 0;
}