Digital DP HDU3652

B-number

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): Accepted submission (s): 2866

Problem Descriptiona Wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the SUB-STR ing "all" and can be divided by 13. For example, and 2613 is wqb-numbers, but 143 and 2639 is not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.

Inputprocess till EOF. There is one positive integer n (1 <= n <= 1000000000).

Outputprint each answer in a single line.

Sample Input131002001000

Sample Output1122

Authorwqb0039

Source of Asia regional Chengdu site--online Contest test Instructions:
Calculates the number of numbers within N that contain 13 and can be divisible by 13:

1 /*2 Memory Search + digital DP, not very understanding of this routine, Dp[i][j][k],i represents the number of bits, j for the remainder, k=0 that no 13,k=1 indicates that the end is 1,3 K=0 says that there are 13. A number divided by 13 can be the remainder of the first divided by 13 and then divided by the last few.4 */5#include <iostream>6#include <string>7#include <cstdio>8#include <cmath>9#include <cstring>Ten#include <algorithm> One#include <vector> A#include <iomanip> -#include <queue> -#include <stack> the using namespacestd; - intN; - intdp[ -][ -][3];  - intc[ -]; + intDfsintLne,intMoDintHaveintLim//whether Lim represents an upper limit - { +     if(lne<=0)//There is no number of digits to return to the condition of compliance A     returnmod==0&have==2; at     if(!lim&&dp[lne][mod][have]!=-1)//there is no limit and has been accessed -     returnDp[lne][mod][have]; -     intNum=lim?c[lne]:9;//assuming that the bit is 2, the next one is 3, and if the bit is 1 now, then the next one is able to fetch 9, -                            //if the bit is 2, the next one can only fetch 3 -     intans=0; -      for(intI=0; i<=num;i++) in     { -         intNmod= (mod*Ten+i)% -;//See if we can divide the 13 evenly, and because it starts with the highest digit of the original number, to                                   //in fact, this process is a division process . +         intNhave=have ; -         if(have==0&&i==1) nhave=1;//The end is not 1, and now it's 1. the         if(have==1&&i!=1&&i!=3) nhave=0;//The end is 1, and now it's not 1. *         if(have==1&&i==3) nhave=2;//The end is 1, and now it's 3. $Ans+=dfs (lne-1, Nmod,nhave,lim&&i==num);//Lim&&i==num, at the very beginning, the NUM removed was the highest,Panax Notoginseng     //So if I is smaller than num, then the next one of I can reach 9, and I==num, the biggest can reach only, bit[pos-1] -     } the     if(!Lim) +Dp[lne][mod][have]=ans;//DP only records no finite values A     returnans; the } + intMain () - { $      while(SCANF ("%d", &n)! =EOF) $     { -memset (dp,-1,sizeof(DP)); -         intCnt=0; the          while(n) -         {Wuyic[++cnt]=n%Ten; theN/=Ten; -         } Wuprintf"%d\n", DFS (CNT,0,0,1)); -     } About     return 0; $}

Digital DP HDU3652