POJ 1953 World Cup Noise

World Cup Noise

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 14397		Accepted: 7129

Description

Background 
"KO-RE-A, KO-RE-A" shout 54.000 happy football fans after their team has reached the semifinals of the FIFA World Cup in their home country. But although their excitement is real, the Korean people are still very organized by nature. For example, they have organized huge trumpets (that sound like blowing a ship's horn) to support their team playing on the field. The fans want to keep the level of noise constant throughout the match. 
The trumpets are operated by compressed gas. However, if you blow the trumpet for 2 seconds without stopping it will break. So when the trumpet makes noise, everything is okay, but in a pause of the trumpet,the fans must chant "KO-RE-A"! 
Before the match, a group of fans gathers and decides on a chanting pattern. The pattern is a sequence of 0's and 1's which is interpreted in the following way: If the pattern shows a 1, the trumpet is blown. If it shows a 0, the fans chant "KO-RE-A". To ensure that the trumpet will not break, the pattern is not allowed to have two consecutive 1's in it. 
Problem 
Given a positive integer n, determine the number of different chanting patterns of this length, i.e., determine the number of n-bit sequences that contain no adjacent 1's. For example, for n = 3 the answer is 5 (sequences 000, 001, 010, 100, 101 are acceptable while 011, 110, 111 are not).

Input

The first line contains the number of scenarios. 
For each scenario, you are given a single positive integer less than 45 on a line by itself.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the number of n-bit sequences which have no adjacent 1's. Terminate the output for the scenario with a blank line.

Sample Input

2
3
1

Sample Output

Scenario #1:
5

Scenario #2:
2
题目大意：求一个长度为n的由0和1组成的序列中满足没有两个1相邻的序列的数目。

解题方法：用动态规划可以很简单的解答出本题，dp方程为dp[i] = dp[i - 1] + dp[i - 2]，一开始我也不明白这道题为什么是这样解答的，其实思想是这样子的，当一个长度为n - 1的01串变为长度为n的01串的时候，在后面添加一个0是没有问题的，添加一个1的组合数其实就是长度为n - 1的01串的组合数，而在后面添加一个1则必须要求长度为n - 1的01串最后一位必须为0，组合数和长度为n - 2的01串是一样的，所以dp方程为dp[i] = dp[i - 1] + dp[i - 2]。

#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    int dp[100] = {0, 2, 3, 5};
    int n, x, nCase;
    for (int i = 4; i <= 45; i++)
    {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    scanf("%d", &n);
    nCase = 0;
    while(n--)
    {
        scanf("%d", &x);
        printf("Scenario #%d:
%d

", ++nCase, dp[x]);
    }
    return 0;
}