数位DP- [hdu 3652] B-number

转载

[hdu 3652] B-number

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

转载地址

题意：找出1~n范围内含有13并且能被13整除的数字的个数

思路：使用记忆化深搜来记录状态，配合数位DP来解决

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int bit[15];
int dp[15][15][3];
//dp[i][j][k]
//i:数位
//j:余数
//k:3种操作状况，0：末尾不是1,1：末尾是1,2：含有13

int dfs(int pos,int mod,int have,int lim)//lim记录上限
{
    int num,i,ans,mod_x,have_x;
    if(pos<=0)
        return mod == 0 && have == 2;
    if(!lim && dp[pos][mod][have] != -1)//没有上限并且已被访问过
        return dp[pos][mod][have];
    num = lim?bit[pos]: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
    ans = 0;
    for(i = 0; i<=num; i++)
    {
        mod_x = (mod*10+i)%13;//看是否能整除13，而且由于是从原来数字最高位开始算，细心的同学可以发现，事实上这个过程就是一个除法过程
        have_x = have;
        if(have == 0 && i == 1)//末尾不是1，现在加入的是1
            have_x = 1;//标记为末尾是1
        if(have == 1 && i != 1)//末尾是1，现在加入的不是1
            have_x = 0;//标记为末尾不是1
        if(have == 1 && i == 3)//末尾是1，现在加入的是3
            have_x = 2;//标记为含有13
        ans+=dfs(pos-1,mod_x,have_x,lim&&i==num);//lim&&i==num，在最开始，取出的num是最高位，所以如果i比num小，那么i的下一位都可以到达9，而i==num了，最大能到达的就只有,bit[pos-1]
    }
    if(!lim)
        dp[pos][mod][have] = ans;
    return ans;
}

int main()
{
    int n,len;
    while(~scanf("%d",&n))
    {
        memset(bit,0,sizeof(bit));
        memset(dp,-1,sizeof(dp));
        len = 0;
        while(n)
        {
            bit[++len] = n%10;
            n/=10;
        }
        printf("%d\n",dfs(len,0,0,1));
    }

    return 0;
}

    
    
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57

自己AC code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[23][13][3];//需要3个状态量才能确定唯一一个状态 ,去掉中间的状态量res，会出现重复，打表观察 
int d[15];

int dfs(int pos,int res,int have,bool lim){
    if(pos == 0){
    //  if(res == 0 && have == 2){
    //      cout << "nihao" << endl;
    //  }
        return res == 0 && have == 2;
    }

    if(!lim && dp[pos][res][have] != -1){
        return dp[pos][res][have];
    }

    int temp = 0;
    int up = lim ? d[pos] : 9;
    for(int i = 0;i <= up;++i){
        int res_t = (res*10 + i) % 13,have_t = have;
        if(have == 0 && i== 1){
            have_t = 1;
        }
        if(have == 1 && i == 3){
            have_t = 2;
        //  cout << "en " << pos-1 << " " << res_t  << " " << have_t<< endl;
        }
        if(have == 1 && i != 1 && i != 3){//
            have_t = 0;
        }
        temp += dfs(pos - 1,res_t,have_t,lim && i == d[pos]);
    }
    if(!lim){
        dp[pos][res][have] = temp;
    }
    return temp;
}
int solve(int x){
    int temp = x;
    int cnt = 0;
    while(temp > 0){
        d[++cnt] = temp % 10;
        temp /= 10;
    }
    d[cnt+1] = 0;
    return dfs(cnt,0,0,true);
}
void init(){
    memset(dp,-1,sizeof(dp));
}
int main(){
    int n = 0;
    while(cin >> n){
        init();
        cout << solve(n) << endl;   
    }

    return 0;
}