Codeforces 55D Beautiful Numbers

Title Link: Http://codeforces.com/problemset/problem/55/D

Test instructions: If a number can be divisible by each of her non-0 bits, then this number is called beautiful numbers.

Given L and R, the total number of beautiful numbers in the Q interval [l,r].

Ideas:

Test instructions is the number that can be divisible by each of its non-0-bit least common multiple.

Because every bit is 1,2,3,4,5,6,7,8,9. That is, the worst case of this number can be divisible by 1*2*3*4*5*6*7*8*9 = 2520.

Set this number to X, set x = 2520*t+y. The LCM is the least common multiple of the number of occurrences of each non-0-bit X.

Then you can know 2520% LCM = 0.

So the x LCM = X%2520%LCM. So in the memory of the search, just save the current number of the remainder of the%2520.

So need a DP[CUR][LAST][LCM] size has 19*2520*2520 to hyper-memory, so the recording of the LCM to disperse a bit.

1#include <bits/stdc++.h>2 using namespacestd;3 Long LongL, R;4 intnum[ A];5 inthash[2550];6 Long Longdp[ A][2530][ -];7 voidInit ()8 {9Memset (DP,-1,sizeof(DP));Ten     intCNT =0; Onehash[0] =0; A      for(inti =1; I <=2520; i++) -     { -         if(2520%i = =0) Hash[i] = + +CNT; the     } - } -  - intgcdintAintb) + { -     if(A < b)returngcd (b, a); +     if(b = =0)returnA; A     returnGCD (b, a%b); at } - intGETLCM (intAintb) - { -     returnA/GCD (A, b) *b; - } - Long LongDfsintCurintLastintLcmintlimit) in { -     if(Cur <0) to     { +         if(last% LCM = =0)return 1; -         Else return 0; the     } *     if(!limit && DP[CUR][LAST][HASH[LCM]]! =-1)returnDP[CUR][LAST][HASH[LCM]]; $     Panax Notoginseng     Long LongRET =0; -     intup = Limit?num[cur]:9; the      for(inti =0; I <= up; i++) +     { A         inttemp = (last*Ten+i)%2520; the         intNOWLCM; +         if(I! =0) NOWLCM =GETLCM (LCM, i); -         ElseNOWLCM =LCM; $RET + = DFS (cur-1, temp, NOWLCM, limit && i = =Up ); $     } -     if(!limit) DP[CUR][LAST][HASH[LCM]] =ret; -     returnret; the } - Long LongSlove (Long Longx)Wuyi { the     intCNT =0; -      while(x) Wu     { -num[cnt++] = x%Ten; AboutX/=Ten; $     } -     returnDFS (cnt-1,0,1,1); - } - intT; A intMain () + { the init (); -scanf"%d", &T); $      while(t--) the     { theCin>>l>>R; theCout<<slove (R)-Slove (L-1) <<Endl; the     } -     return 0; in}

Codeforces 55D Beautiful Numbers