【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
 
 
【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
 
【分析】
 

  f[i][j][k]表示填i个数,模13余数为j,状态为k的方案数。
  k=0表示没有‘13’且末位不为1,k=1表示没有‘13’但末位为1,k=3为含有‘13’。

 

  这是数位DP模版题啊。。感觉我真的从难往简单做了,第一题搞的我,这酸爽!!

  我还是too naive啊,一开始弄k表示状态想得超复杂,觉得要记录含不含k以及上一位是什么,其实很多状态是没用的嘛,只要知道上一位是不是1就好了,我是如此搞笑!

 

代码如下:

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<cmath>
 8 using namespace std;
 9 #define LL long long
10 
11 int f[20][20][3],a[20];
12 int c;
13 
14 LL ffind(int n,int k,int tt,bool flag)
15 {
16     if(n==0) return (tt==2&&k==0);
17     if(!flag&&f[n][k][tt]!=-1) return f[n][k][tt];
18     LL ans=0;
19     int ed=flag?a[n]:9;
20     for(int i=0;i<=ed;i++)
21     {
22         int nt=0;
23         if(i==1) nt=1;
24         if(tt==1&&i==3) nt=2;
25         if(tt==2) nt=2;
26         ans+=ffind(n-1,(k*10+i)%13,nt,(flag&&(i==ed)));
27     }
28     if(!flag) f[n][k][tt]=ans;
29     return ans;
30 }
31 
32 LL get_ans(int n)
33 {
34     int x=n;
35     c=0;
36     while(x) a[++c]=x%10,x/=10;
37     return ffind(c,0,0,1);
38 }
39 
40 int main()
41 {
42     memset(f,-1,sizeof(f));
43     int n;
44     while(scanf("%d",&n)!=EOF)
45     {
46         LL ans=get_ans(n);
47         printf("%lld\n",ans);
48     }
49     return 0;
50 }
[HDU 3652]

 

记得100年前,没做过几道数位DP的我比赛时打了个高精数位DP那么勇猛,还AC了,现在真是返老还童了ORZ!!

 

2016-10-08 21:28:51

  
posted @ 2016-10-08 21:24  konjak魔芋  阅读(186)  评论(0编辑  收藏  举报