Count 101 hdu 3485

Count 101

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 373    Accepted Submission(s): 196

Problem 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 
    
    
    
    
     
   
   
   
   

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

找到规律：

f[i]=f[i-1]+f[i-2]+f[i-4];

  #include<iostream> using namespace std; int f[10005]; int n; int deal(int n) { if(f[n]!=0) return f[n]; f[n-1]=deal(n-1)%9997; f[n-2]=deal(n-2)%9997; f[n-4]=deal(n-4)%9997; return ((f[n-1]+f[n-2]+f[n-4])%9997); } int main() { while(scanf("%d",&n)!=EOF&&n!=-1) { f[0]=1; f[1]=2; f[2]=4; f[3]=7; printf("%d/n",deal(n)); } }