HDU 3652 B-number (数位dp)

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7409    Accepted Submission(s): 4340

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

131002001000

 

Sample Output

1122

 

#include <stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[20];int dp[20][20][3];//sta 为判断余数是否为0   sta1 0 不为1  1为1 2为 13 int dfs(int pos,int mod,int have,bool limit)//sta 为 mod sta1为数位判断 {if(pos==-1)return mod==0&&have==2;if(dp[pos][mod][have]!=-1&&!limit)return dp[pos][mod][have];int up=limit?a[pos]:9;int sum=0;int have_x,i;for(i=0;i<=up;i++){have_x=have;if(have_x==0&&i==1){have_x=1;}if(have_x==1&&i==3){have_x=2;}  else if(have_x==1&&i!=1){have_x=0;}sum+=dfs(pos-1,(mod*10+i)%13,have_x,limit&&i==up);}if(!limit){dp[pos][mod][have]=sum;}return sum;}int solve(int x){int cnt=0;while(x){a[cnt++]=x%10;x/=10;}return dfs(cnt-1,0,0,1);}int main(int argc, char *argv[]){memset(dp,-1,sizeof(dp));memset(a,-1,sizeof(a));int n;while(scanf("%d",&n)!=EOF){printf("%d\n",solve(n));}return 0;}