poj1953

Posted 2022-04-01 kasenbob

1．链接地址

 https://vjudge.net/problem/POJ-1953

2．问题描述

 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).

输入样例

2
3
1

输出样例

Scenario #1:
5

Scenario #2:
2

3．解题思路

 假设有n个空要填，队列称为f（n），第一个填了1，第二个只能填0，后面就是f（n-2）的排列；假设第一个填了0，后面就可以是f（n-1）的排列，所以这其实是一个斐波那契数列

4．算法实现源代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 50;
long long int arry[maxn];

void init()

    memset(arry,0,sizeof(arry));
    arry[1]=2;
    arry[2]=3;
    for(int i=3;i<50;i++)
    
        arry[i]=arry[i-1]+arry[i-2];
    


int main()

    init(); 
    int n;
    scanf("%d",&n);
    int ans,temp=1;
    for(int i=0;i<n;i++)
    
        scanf("%d",&ans);
        printf("Scenario #%d:\n%lld\n\n",temp,arry[ans]);
        temp++;