[HDOJ3652]B-Number（数位dp）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3652

题意：找区间内的数，数内含有连续的13并且这个数能被13整除。

dfs(int l, bool one, bool three, int sum, bool flag, bool ok),分别记录位数，上一位1是否出现，当前位3是否出现，这个数%13的值，是否到达边界以及这条dfs链上的结果是否已经有13了。状态设计得有点傻B了。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define fr first
 4 #define sc second
 5 #define cl clear
 6 #define BUG puts("here!!!")
 7 #define W(a) while(a--)
 8 #define pb(a) push_back(a)
 9 #define Rint(a) scanf("%d", &a)
10 #define Rll(a) scanf("%I64d", &a)
11 #define Rs(a) scanf("%s", a)
12 #define Cin(a) cin >> a
13 #define FRead() freopen("in", "r", stdin)
14 #define FWrite() freopen("out", "w", stdout)
15 #define Rep(i, len) for(int i = 0; i < (len); i++)
16 #define For(i, a, len) for(int i = (a); i < (len); i++)
17 #define Cls(a) memset((a), 0, sizeof(a))
18 #define Clr(a, x) memset((a), (x), sizeof(a))
19 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
20 #define lrt rt << 1
21 #define rrt rt << 1 | 1
22 #define pi 3.14159265359
23 #define RT return
24 #define lowbit(x) x & (-x)
25 #define onecnt(x) __builtin_popcount(x)
26 typedef long long LL;
27 typedef long double LD;
28 typedef unsigned long long ULL;
29 typedef pair<int, int> pii;
30 typedef pair<string, int> psi;
31 typedef pair<LL, LL> pll;
32 typedef map<string, int> msi;
33 typedef vector<int> vi;
34 typedef vector<LL> vl;
35 typedef vector<vl> vvl;
36 typedef vector<bool> vb;
37
38 const int maxn = 15;
39 int digit[maxn];
40 LL dp[maxn][2][2][15][2];
41 LL n;
42
43 LL dfs(int l, bool one, bool three, int sum, bool flag, bool ok) {
44   if(l == 0) {
45     if(sum != 0) return 0;
46     if(ok) return 1;
47     return 0;
48   }
49   if(!flag && ~dp[l][one][three][sum][ok]) return dp[l][one][three][sum][ok];
50   LL ret = 0;
51   int pos = flag ? digit[l] : 9;
52   Rep(i, pos+1) {
53     if(one && i == 3) ret += dfs(l-1, one, true, (sum*10+i)%13, flag&&(i==pos), true);
54     else if(i == 1) ret += dfs(l-1, true, false, (sum*10+i)%13, flag&&(i==pos), ok);
55     else ret += dfs(l-1, false, false, (sum*10+i)%13, flag&&(i==pos), ok);
56   }
57   if(!flag) dp[l][one][three][sum][ok] = ret;
58   return ret;
59 }
60
61 LL f(LL x) {
62   int pos = 0;
63   while(x) {
64     digit[++pos] = x % 10;
65     x /= 10;
66   }
67   return dfs(pos, false, false, 0, true, false);
68 }
69
70 signed main() {
71   //FRead();
72   Clr(dp, -1);
73   while(cin >> n) {
74     cout << f(n) << endl;
75   }
76   RT 0;
77 }