Timus 1225 flags basic DP simple recursion

1225. FlagsTime limit:1.0 Second
Memory limit:64 MBOn the day of the "the Flag of Russia a shop-owner decided to decorate the show-window of the" with textile stripes of White, blue and red colors. He wants to satisfy the following conditions:

1. Stripes of the same color cannot is placed next to all other.
2. A blue stripe must always being 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:Input n, the number of the stripes, 1≤ N ≤45.Output M, the number of the ways to decorate the Shop-window.Sample

input	Output
3	4

1 Stands for White

2. Blue

3. Red

DP[1][J] is selected to the first I stripe, the first I is J this color

and then recursion.

Attention:

1. To use a long long

2. Blue can not be placed on the last and the front, only in the middle

1#include <iostream>2#include <cstring>3 4 using namespacestd;5 6 Const intmaxn= -;7 8 Long Longdp[maxn][4];9 Ten intMain () One { Adp[0][1]=dp[0][2]=dp[0][3]=0; -dp[1][1]=dp[1][3]=1; -dp[1][2]=0; the  -      for(intI=2; i<= $; i++) -     { -dp[i][1]=dp[i-1][3]+dp[i-2][3]; +dp[i][2]=dp[i-1][1]+dp[i-1][3]; -dp[i][3]=dp[i-1][1]+dp[i-2][1]; +     } A  at     intN; -  -      while(cin>>N) -     { -         Long Longsum=dp[n][1]+dp[n][3]; -  incout<<sum<<Endl; -     } to     return 0; +}

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Timus 1225 flags basic DP simple recursion