数位DP。。。。
Count 101
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 912 Accepted Submission(s): 477
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 HintWe 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
#include#include #include #include using namespace std; const int mod=9997; int dp[11000][2][2],n; int ans[11000]; void init() { dp[1][1][0]=dp[1][0][0]=1; dp[2][0][0]=dp[2][1][0]=1; dp[2][0][1]=dp[2][1][1]=1; ans[0]=0;ans[1]=2;ans[2]=4; for(int i=3;i<11000;i++) { for(int j=0;j<2;j++) { for(int k=0;k<2;k++) { if(j==0) { dp[i][j][k]+=(dp[i-1][k][0]+dp[i-1][k][1])%mod; } else if(j==1) { if(k==1) { dp[i][j][k]+=(dp[i-1][k][0]+dp[i-1][k][1])%mod; } else { dp[i][j][k]+=dp[i-1][k][0]; } } dp[i][j][k]=dp[i][j][k]%mod; } } ans[i]=(dp[i][1][0]+dp[i][1][1]+dp[i][0][1]+dp[i][0][0])%mod; } } int main() { init(); while(scanf("%d",&n)!=EOF&&~n) { printf("%d\n",ans[n]); } return 0; }