数位dp——hud3652

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

模板与前一篇博客是一样的，只是dp数组的意义略微做了改动。

f[pos][p][is13]表示到第pos位，除以13的余数为p，是否出现过13的数字个数。

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 int f[][][],bit[],n,l;
 int get(int a,int b){
     if(b==)return ;
     return a==?:a==&&b==?:;
 }
 int dfs(int pos,int lim,int p,int is13){
     if(pos<=&&p==&&is13==)return ;
     if(pos<=)return ;
     if(!lim&&f[pos][p][is13]!=-)return f[pos][p][is13];
     int rng=(lim?bit[pos]:),ret=;
     for(int i=;i<=rng;i++)
         ret+=dfs(pos-,lim&&(i==rng),(p*+i)%,get(i,is13));
     if(!lim)f[pos][p][is13]=ret;
     return ret;
 }

 int main(){
     while(scanf("%d",&n)!=EOF){
         for(l=;n;)bit[++l]=n%,n/=;
         memset(f,-,sizeof(f));
         printf("%d\n",dfs(l,,,));
     }
 }