HDU 3485 Count 101（递推）

C - Count 101

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3485

Description

You know YaoYao is fond of his chains. He has a lot of chains and each chain has n diamonds on it. There are two kinds of diamonds, labeled 0 and 1. We can write down the label of diamonds on a chain. So each chain can be written as a sequence consisting of 0 and 1. 
We know that chains are different with each other. And their length is exactly n. And what’s more, each chain sequence doesn’t contain “101” as a substring. 
Could you tell how many chains will YaoYao have at most? 

 

Input

There will be multiple test cases in a test data. For each test case, there is only one number n(n<10000). The end of the input is indicated by a -1, which should not be processed as a case.

 

Output

For each test case, only one line with a number indicating the total number of chains YaoYao can have at most of length n. The answer should be print after module 9997.

 

Sample Input

3 4 -1

 

Sample Output

7 12

Hint

We can see when the length equals to 4. We can have those chains: 0000,0001,0010,0011 0100,0110,0111,1000 1001,1100,1110,1111 


有一串由n个珠子连成的链 每个珠子的标号只能是 0或1 求总共有多少种不同的排列

一开始还在想是不是要用到位运算 没想到竟然是递推= =

#include<cstdio>
#include<iostream>
#include<bitset>
#include<algorithm>
using namespace std;
int a[10000+10];
int main()
{
    int n;
    int i,j;
    a[0]=1;
    a[1]=2;
    a[2]=4;
    a[3]=7;
    a[4]=12;
    for(i=5;i<=10000;i++)
    {
        a[i]=(a[i-1]+a[i-2]+a[i-4])%9997;
    }
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1) break;
        printf("%d
",a[n]);
    }
    return 0;
}