hdu3652 B-number（数位dp+dfs）

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3376 Accepted Submission(s): 1891

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 

题意：找出1~n有多少个数既含有13又能被13整除。

分析：记忆化搜索配合数位dp求解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int dp[15][15][3],s[15];//dp[i][j][k]，i表示位数，j表示余数，k表示末尾是1、末尾不是1、含有13.

int dfs(int pos, int mod, int have, int lim)//前三个数对应数组dp，lim表示上限
{
    int num,ans,mod_x,have_x;
    if (pos <= 0)
        return mod==0&&have==2;
    if (!lim && dp[pos][mod][have]!=-1) //没有上限且被访问过
        return dp[pos][mod][have];
    ans = 0;
    num = lim?s[pos]:9;//如果有上限，只能取到当前位数，如果没上限，可取到9
                        //假设该位是2，下一位是3，如果现在算到该位为1，那么下一位是能取到9的，如果该位为2，下一位只能取到3
    for (int i=0; i<=num; i++)
    {
        mod_x = (mod*10+i)%13;//该位的每种情况对13取模
        have_x = have;
        if (have==0 && i==1)//末尾加1
            have_x=1;
        if (have==1 && i!=1)//末尾已经为1了
            have_x=0;
        if (have==1 && i==3)//末尾是1，现在加3
            have_x=2;
        ans+=dfs(pos-1, mod_x, have_x, lim&&i==num);//如果i==num，下一位能取的最大数就为s[pos-1]，i！=num，下一位能取到9
    }
    if (!lim)
        dp[pos][mod][have] = ans;
    return ans;
}

int main ()
{
    int n,t;
    while (scanf ("%d",&n)==1)
    {
        CL(s);
        memset(dp, -1, sizeof(dp));
        t = 0;
        while (n)
        {
            s[++t]=n%10;
            n/=10;
        }
        cout<<dfs(t, 0, 0, 1)<<endl;
    }
    return 0;
}