ural1225-Flags-动态规划

1225. Flags

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:

1. Stripes of the same color cannot be placed next to each other.
2. 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:

Input

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

Output

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

Sample

inputoutput

3

4

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

Tags: dynamic programming  (

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Difficulty: 56    Printable version    Submit solution    Discussion (17)
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我太弱了，看了如此多稀奇古怪说不清楚的题解。。。。

还是自己想出来了，尽管题目很简单。。。

但好像我这个说的清楚一些

//该题//1.注意long long啊！斐波那契数列到第45个正好爆int！//2.dp[i]=dp[i-1]+dp[i-2]这个方程是这样得到的！//原因：①当n=i-1时，最后一块旗子的颜色不是蓝色//②所以当n=i时，最后一块旗子的颜色也不是蓝色，且若从n=i-1时得到该旗子，则最后一面旗子与n=i-1的最后一面颜色也不一样//因而dp[i]可由dp[i-1]得到//③当n=i时，倒数第二块旗子的颜色为蓝（即与②中的倒数第二块为红白颜色不一样），此时对最后一面旗子与倒数第三面旗子一样,//因而dp[i]可由dp[i-2]得到#include <iostream>#include <cstdio>using namespace std;const int maxnum=45;typedef long long ll;ll dp[maxnum]={2,2,4};int main(){    int n;    cin>>n;    for(int i=3;i<45;++i)        dp[i]=dp[i-1]+dp[i-2];    cout<<dp[n-1];    return 0;}