poj1953

World Cup Noise

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 17716 Accepted: 8634

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

231

Sample Output

Scenario #1:5Scenario #2:2

Source

TUD Programming Contest 2002, Darmstadt, Germany

题目大意：现在给一个n，代表有n个数，这n个数只能取0或1，要求不能出现两个连续的1，求一共有多少种情况。

题目分析：其实这道题不用分析，看数据也能得出这就是一个斐波拉切数列，只不过首项不同.对于每个n串有两种情况，一种是在n-1串后加0，或者在n-2串后加01.

代码:

#include <iostream>using namespace std;int main(){    int T,k=1;    int dp[50];    cin>>T;    dp[1]=2;    dp[2]=3;    for(int i=3;i<50;i++){        dp[i]=dp[i-1]+dp[i-2];    }    while(T--)    {        int n;        cin>>n;        cout<<"Scenario #"<<k++<<":"<<endl;        cout<<dp[n]<<endl<<endl;    }    return 0;}