2010 ACM-ICPC Multi-University Training Contest(5)HDU3485Count 101题解动态规划DP

Count 101

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 400    Accepted Submission(s): 219

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

 

 

Recommend

zhengfeng

状态：

d[i][j][k]表示前i个数第i-1个是j第i-2个是k的满足条件的串的个数

状态转移方程：

d[i][k][0]=(d[i][k][0]+d[i-1][j][k])%9997
 d[i][k][1]=(d[i][k][1]+d[i-1][j][k])%9997 (jk!='10')

边界：

d[1][0][0]=d[1][0][1]=1

d[2][j][k]=1

 

代码：

#include<stdio.h> int main() { int n; while(scanf("%d",&n),n!=-1) { int i,j,k,ans=0,d[10005][2][2]={0}; d[1][0][0]=d[1][0][1]=1; for(j=0;j<2;j++) for(k=0;k<2;k++) d[2][j][k]=1; for(i=3;i<=n+1;i++) for(j=0;j<2;j++) for(k=0;k<2;k++) if(j==1&&k==0) d[i][k][0]=(d[i][k][0]+d[i-1][j][k])%9997; else { d[i][k][0]=(d[i][k][0]+d[i-1][j][k])%9997; d[i][k][1]=(d[i][k][1]+d[i-1][j][k])%9997; } for(j=0;j<2;j++) for(k=0;k<2;k++) ans=(ans+d[n][j][k])%9997;//刚开始忘了取模，不停WA printf("%d/n",ans); } }