Flags-Ural1225简单递推

Time limit: 1.0 second Memory limit: 64 MB

On the Day of the Flag of Russia a shop-owner decided to decorate the show-window of his shop with textile stripes of white, blue and red colors. He wants to satisfy the following conditions:
Stripes of the same color cannot be placed next to each other.
A blue stripe must always be placed between a white and a red or between a red and a white one.
Determine the number of the ways to fulfill his wish.
Example. For N = 3 result is following:
Problem illustration

Input

N, the number of the stripes, 1 ≤ N ≤ 45.

Output

M, the number of the ways to decorate the shop-window.

Sample

input
3
output
4

Problem Source:

2002-2003 ACM Central Region of Russia Quarterfinal Programming Contest, Rybinsk, October 2002

简单的递推，三种颜色的布，相邻的颜色不能相同，蓝色在红色与白色之间，则用1表示红色，2表示白色，Dp[i][j]表示第i个条纹的颜色为j是的状态数目，当j=1时，如果i-1的颜色为白色，则Dp[i][1]+=Dp[i-1][2],如果为蓝色，则取决于i-2的颜色为白色状态，所以Dp[i][1]=Dp[i-1][2]+Dp[i-2][2],则Dp[i][2]=Dp[i-1][1]+Dp[i-2][1].

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <string>#include <queue>#include <vector>#include <deque>#include <iostream>#include <algorithm>using namespace std;long long Dp[50][3];int main(){    memset(Dp,0,sizeof(Dp));    Dp[1][1]=Dp[1][2]=1;    for(int i=2;i<=45;i++)    {        Dp[i][1] =Dp[i-1][2]+Dp[i-2][2];        Dp[i][2] = Dp[i-1][1]+Dp[i-2][1];    }    int n;    while(~scanf("%d",&n))    {        cout<<Dp[n][1]+Dp[n][2]<<endl;    }    return 0;}