HDU 4389 x Mod f (x) (Digital DP)

Finally, I got this question... I watched it for half a week. The number of questions accumulated in the competition is decreasing... Come on!

Digit DP. f [I] [Sum] [mod] [res] indicates the first I bit, sum, Mod, sum % mod, and Res. F [I + 1] [Sum + k] [mod] [(RES * 10 + k) % mod] + = f [I] [Sum] [mod] [res];

All F [I] [Sum] [mod] [res] are processed, and the statistics are calculated from high to low. For example, if the current BIT is a_ I, then the enumerated Current BIT is 0... (a_i-1) these numbers, all the States of the number after the I bit has been preprocessed... So % mod = 0 can be accumulated directly... Finally, the bitwise is processed separately...

1 # include <iostream> 2 # include <cstdio> 3 # include <cmath> 4 # include <vector> 5 # include <cstring> 6 # include <algorithm> 7 # include <string> 8 # include <set> 9 # include <ctime> 10 # include <queue> 11 # include <map> 12 # include <functional> 13 # include <numeric> 14 # include <sstream> 15 16 # define Cl (ARR, val) memset (ARR, Val, sizeof (ARR) 17 # define rep (I, n) for (I) = 0; (I) <(N ); ++ (I) 18 # Define for (I, L, H) for (I) = (l); (I) <= (h); ++ (I )) 19 # define Ford (I, H, L) for (I) = (h); (I)> = (l); -- (I )) 20 # define L (x) <1 21 # define R (x) <1 | 1 22 # define mid (L, R) (L + r)> 1 23 # define min (x, y) x <Y? X: Y 24 # define max (x, y) x <Y? Y: x 25 # define e (x) (1 <(x) 26 27 const int EPS = 1e-4; 28 typedef long ll; 29 const int INF = ~ 0u> 2; 30 using namespace STD; 31 32 const int n = 83; 33 34 int f [12] [N] [N] [N]; 35 int ten [15]; 36 37 38 void Init () {39 int I, sum, Mod, res, K, J; 40 ten [0] = 1; 41 for (I = 1; I <= 10; + I) Ten [I] = ten [I-1] * 10; 42 Cl (F, 0 ); 43 for (I = 0; I <10; ++ I) {44 for (j = 1; j <n; ++ J) f [1] [I] [J] [I % J] = 1; 45} 46 47 for (I = 1; I <9; ++ I) {48 for (sum = 0; sum <n; ++ sum) {49 for (mod = 1; mod <n; + + Mod) {50 for (RES = 0; Res <n; ++ res) {51 if (! F [I] [Sum] [mod] [res]) continue; 52 for (k = 0; k <10 & sum + k <n; ++ K) 53 f [I + 1] [Sum + k] [mod] [(RES * 10 + k) % mod] + = f [I] [Sum] [mod] [res]; 54} 55} 56} 57} 58} 59 60 int A [15]; 61 62 int do (int x) {63 If (x <= 10) return X; 64 int tsum = 0, I, j, n = 0, U; 65 int mod, sum, num, res, ANS = 0; 66 67 for (u = x; u> 0; U/= 10) A [++ N] = u % 10, tsum + = u % 10; 68 69 for (mod = 1; mod <= 9 * n; ++ mod ){/ /Enumeration modulo, that is, all digits and 70 If (mod> X) break; 71 sum = MOD; 72 num = 0; 73 for (I = N; i> 1; -- I) {74 for (j = 0; j <A [I] & J <= sum; ++ J) {// from 0... (a_ I-1) enumeration. 75 for (RES = 0; Res <MOD; ++ res) {// res indicates the number of digits after I bit and the value obtained by modulo 76 if (Num + J * Ten [I-1] + Res) % mod = 0) {// according to the same remainder formula 77 ans + = f [I-1] [Sum-J] [mod] [res]; 78} 79} 80} 81 sum-= A [I]; 82 num + = A [I] * Ten [I-1]; 83} 84} 85 while (1) {// handle the random bit 86 If (X % tsum = 0) ans ++; 87 If (X % 10 = 0) break; 88 X --; tsum --; 89} 90 return ans; 91} 92 93 int main () {94 freopen ("data. in "," r ", stdin); 95 96 Init (); 97 int t, a, B, CAS = 0; 98 scanf (" % d ", & T ); 99 while (t --) {100 scanf ("% d", & A, & B); 101 printf ("case % d: % d \ n ", + + cas, do (B)-Do (A-1); 102} 103 return 0; 104}