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[20][14][4],bit[20],n,l;
int get(int a,int b){
    if(b==2)return 2;
    return a==1?1:a==3&&b==1?2:0;
}
int dfs(int pos,int lim,int p,int is13){
    if(pos<=0&&p==0&&is13==2)return 1;
    if(pos<=0)return 0;
    if(!lim&&f[pos][p][is13]!=-1)return f[pos][p][is13];
    int rng=(lim?bit[pos]:9),ret=0;
    for(int i=0;i<=rng;i++)
        ret+=dfs(pos-1,lim&&(i==rng),(p*10+i)%13,get(i,is13));
    if(!lim)f[pos][p][is13]=ret;
    return ret;
}

int main(){
    while(scanf("%d",&n)!=EOF){
        for(l=0;n;)bit[++l]=n%10,n/=10;
        memset(f,-1,sizeof(f));
        printf("%d
",dfs(l,1,0,0));
    }
}