HDU 3652 B-number （数位DP）

B-number

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

 

Author

wqb0039

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

Recommend

lcy

 

题目大意：

给定一个数n，问你 1~n 有多少个数本身既是13的倍数，且数位中含有13

解题思路：

很明显是数位DP，用 长度，是否包含13，前一个数位，以及对13求余后的余数 这四维来表示DP的状态。

解题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>
using namespace std;

int dp[11][2][15][15];
const int w[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
string st;

int DP(int len,int f,int pre,int left){
    if(len==0){
        if(f==1 && left==0 ) return 1;
        else return 0;
    }
    if(dp[len][f][pre][left]!=-1) return dp[len][f][pre][left];
    int ans=0;
    for(int i=0;i<=9;i++){
        if(i==3 && pre==1 ) ans+=DP(len-1,1,i,(left+w[len-1]*i)%13 );
        else ans+=DP(len-1,f,i,(left+w[len-1]*i)%13 );
    }
    return dp[len][f][pre][left]=ans;
}

int dfs(int len,int f,int pre,int left){
    if(len<=0){
        if(f==1 && left==0 ) return 1;
        else return 0;
    }
    int x=st[st.length()-len]-'0';
    int ans=0;
    for(int i=0;i<x;i++){
        if(pre==1 && i==3) ans+=DP(len-1,1,i,( left+ i*w[len-1])%13 );
        else ans+=DP(len-1,f,i,( left+ i*w[len-1])%13 );
    }
    if(pre==1 && x==3) ans+=dfs(len-1,1,x, ( left+ x*w[len-1])%13 );
    else  ans+=dfs(len-1,f,x, ( left+ x*w[len-1])%13 );
    return ans;
}

int main(){
    int n;
    memset(dp,-1,sizeof(dp));
    while(cin>>n){
        stringstream ss;
        ss<<n;
        ss>>st;
        cout<<dfs(st.length(),0,0,0)<<endl;
    }
    return 0;
}