hdu3652——B-number（数位dp变形）

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

加了一个还要被13整除的条件，那就在状态数组多加一项表示被13整除的情况，当这项等于0说明能被13整除

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 10000000005#define Mod 10001using namespace std;int dight[30],dp[30][3][13];int dfs(int pos,int s,bool limit,int mod){    if(pos==0)        return s==2&&mod==0;    if(!limit&&dp[pos][s][mod]!=-1)        return dp[pos][s][mod];    int end,ret=0;    if(limit)        end=dight[pos];    else        end=9;    for(int d=0;d<=end;++d)    {        int have=s,nmod=(mod*10+d)%13;        if(s==1&&d==3)            have=2;        if(s==0&&d==1)            have=1;        if(s==1&&d!=1&&d!=3)            have=0;        ret+=dfs(pos-1,have,limit&&d==end,nmod);    }    if(!limit)        dp[pos][s][mod]=ret;    return ret;}int solve(int a){    memset(dight,0,sizeof(dight));    int cnt=1;    while(a!=0)    {        dight[cnt++]=a%10;        a/=10;    }    return dfs(cnt-1,0,1,0);}int main(){    memset(dp,-1,sizeof(dp));    int n;    while(~scanf("%d",&n))    {        printf("%d\n",solve(n));    }    return 0;}