hdu 3652 B-number

B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1654 Accepted Submission(s): 902


Problem Description
A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.


Input
Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).


Output
Print each answer in a single line.


Sample Input
13
100
200
1000


Sample Output
1
1
2
2


Author
wqb0039


Source
2010 Asia Regional Chengdu Site —— Online Contest


Recommend
lcy

 

参考： http://hi.baidu.com/sunnydaughter/item/69d3f9330b7c7f88c2cf29ea

 

 1 //31MS    272K    943 B    C++
 2 //记忆化搜索解数位DP
 3 /*
 4 
 5 dp[i][j][k]：
 6    i表示数的第i位
 7    j表示第i位前的位数和 %13 
 8    k表示是否有13(0为没有,1为有'1',2为有'13')
 9     
10 */ 
11 #include<stdio.h>
12 #include<string.h>
13 int dp[10][15][3];
14 int digit[10];
15 int dfs(int pos,int pre,int have,int judge)
16 {
17     if(pos==-1) return pre==0&&have==2; //模13为0且有'13'符合 
18     if(!judge && dp[pos][pre][have]!=-1)
19         return dp[pos][pre][have];
20     int ans=0;
21     int end=judge?digit[pos]:9;
22     for(int i=0;i<=end;i++){
23         int tpre=(pre*10+i)%13;
24         int thave=have;
25         if(have==1 && i!=1) thave=0;
26         if(have==0 && i==1) thave=1;
27         if(have==1 && i==3) thave=2;
28         ans+=dfs(pos-1,tpre,thave,judge&&i==end);
29     }
30     if(!judge) dp[pos][pre][have]=ans;
31     return ans;  
32 }
33 int deal(int n)
34 {
35     memset(digit,0,sizeof(digit));
36     memset(dp,-1,sizeof(dp));
37     int pos=0;
38     while(n){
39         digit[pos++]=n%10;
40         n/=10;
41     }
42     return dfs(pos-1,0,0,1);
43 }
44 int main(void)
45 {
46     int n;
47     while(scanf("%d",&n)!=EOF)
48     {
49         printf("%d\n",deal(n));
50     }
51     return 0;
52 }

 

posted @ 2013-10-05 15:23  heaventouch  阅读(162)  评论(0)  编辑  收藏  举报