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
100
200
1000
Sample Output
1
1
2
2
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,,,));
}
}