hdu3652 B-number[数位dp] B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6020 Accepted Submission(s): 3473

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

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=20;
ll n,f[N][13][3];int T,bits[N];
ll dfs(int pos,int mod,int st,bool lim){
    if(!pos) return st==2&&!mod;
    ll &res=f[pos][mod][st],ans=0;
    if(!lim&&(~res)) return res;
    int up=!lim?9:bits[pos];
    for(int i=up;~i;i--){
        int p=(mod*10+i)%13;
        if(st==2||(st==1&&i==3))
            ans+=dfs(pos-1,p,2,lim&&i==bits[pos]);
        else if(i==1)
            ans+=dfs(pos-1,p,1,lim&&i==bits[pos]);
        else 
            ans+=dfs(pos-1,p,0,lim&&i==bits[pos]);
    }
    if(!lim) res=ans;
    return ans;
}
ll solve(ll x){
    int len=0;
    for(;x;x/=10) bits[++len]=x%10;
    return dfs(len,0,0,1);
}
int main(){
    memset(f,-1,sizeof f);
    while(~scanf("%I64d",&n))
        printf("%I64d
",solve(n));
    return 0;
}

hdu3652 B-number[数位dp] B-number

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