10.06 tolerance Practice T3 simple tolerance + go to re-search

Description~cirno found a Baka number, this number ~ only 2 and ⑨ two kinds of numbers ~ ~
Now Cirno want to know ~ an interval ~ ~ How many numbers can be divisible by Baka number ~
But Cirno is so talented that the goblins don't bother to count.
You can only rely on the smart you. Input line integer L R (1 < L < R < 10^10) output a positive integer representing the answer to the sample INPUT1 100Sample Output58 The lucky number of the previous scoi is actually a good thing, that is, preprocessing Out all the 2,9-related numbers, and then the numbers go back in and run (if it's pretty scary, 18000+ 446) Search Search is very common, pay attention to the next shift is LCM, others see lucky Number (once on the very rare) code:

1#include <iostream>2#include <cstdio>3#include <algorithm>4 #defineN 1000055 using namespacestd;6 Long LongAns,a[n],l,r,tot,check[n];7 voidPreLong LongDepLong Longval) {8 //cout<<val<<endl;9 //System ("pause");Ten     if(dep==9){ One         if(!val)return; Aa[++tot]=Val; -         return; -     } thePre (dep+1, val*Ten+2); -Pre (dep+1, val*Ten+9); -Pre (dep+1, Val); - } + voidUni () { -Sort (A +1, a+tot+1); +Tot=unique (A +1, a+tot+1)-a-1; A     Long Longtemp=0; at     Long Longb[20000]; -      for(Long LongI=1; i<=tot;i++) b[i]=A[i]; -      for(Long LongI=1; i<=tot;i++){ -         if(Check[i])Continue; -a[++temp]=B[i]; -          for(Long Longj=i+1; j<=tot;j++){ in             if(B[j]%b[i])Continue; -check[j]=1; to         }     +     } -tot=temp; the     return; * } $ Long LonggcdLong LongALong Longb) {Panax Notoginseng     if(a<b) Swap ( A, a); -      while(a=a%b) Swap ( A, a); the     returnb; + } A Long LongLcmLong LongALong Longb) { the     returnA/GCD (A, b) *b; + } - voidDfsLong LongDepLong LongValintNow ) { $     if(Val>r)return; $     if(dep==tot) { -         if(val==1)return; -     //cout<<val<<endl; the         if(now%2==1) -ans+=r/val-(l1)/Val; Wuyi         Else theans-=r/val-(l1)/Val; -         return ; Wu     } -DFS (dep+1, LCM (val,a[dep+1]), now+1); AboutDFS (dep+1, Val,now); $ } - intMain () { -Pre0,0); - Uni (); A //for (int i=1;i<=tot;i++) { + //cout<<a[i]<<endl; the //if (i%100==0) - //System ("pause"); $ //    } theCin>>l>>R; theDfs0,1,0); thecout<<ans; the     return 0; -}

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10.06 tolerance Practice T3 simple tolerance + go to re-search