HDU XXXX: Number of primes [l,r] (DP)

Problem Gtime limit:2000/1000ms (java/other) Memory limit:131072/131072k (Java/other) total submission (s): Acce pted Submission (s): 28Font:Times New Roman |  Verdana | Georgia Font Size: ← → Problem descriptions number is the number which the sum of every digit is a prime number, such as number 98, 29. Output the number ofS number in [l,r]. Input First line containsT(t≤ten) Denoting the number of test cases.
TCases follows for each cases:
There numbers < Span id= "mathjax-span-16" class= "Mrow" >l ,r . (0 ≤l≤ r≤ 1016) Outputfor each case, output the number of S number . Sample Input

24 3049 173

Sample Output

1245

Test instructions: Find out [0,r]-[0,l-1] on the sum of the primes in [L,r].
Idea: Because the data range in 0~10^16 so large, so can not be violent, considering that each bit is up to 9, then a maximum of 15 9 words is 135 number so many, so I made a judgment 135 inside the number of which is the number of the table, and then the digital DP

1#include <cstdio>2#include <iostream>3#include <cstring>4 using namespacestd;5 #defineN 206 intBit[n];7 Long Longdp[n][ $][2];8 intprime[ $];9 Ten BOOLCheckintx) One { A      for(intI=2; i*i<=x;i++){ -         if(x%i==0)return false; -     } the     returnx!=1; - } -  - voidP () + { -      for(intI=2; i<= $; i++){ +         if(check (i)) { Aprime[i]=1; at         } -     } - } - //flag Indicates whether the previous number is the prefix of the upper bound (that is, the following number can be filled in arbitrarily).  - Long LongDfsintPosintStintHaveintflag) - { in     if(!pos)returnhave ; -     if(flag&&dp[pos][st][have]!=-1)returnDp[pos][st][have]; to     Long Longans=0; +     intU=flag?9: Bit[pos]; -      for(intD=0;d <=u;d++){ theAns+=dfs (pos-1, St+d,prime[st+d],flag| | d<u); *         //determines whether the previous position and plus the current position can be a prime number $     }Panax Notoginseng     if(flag) dp[pos][st][have]=ans; -     returnans; the } +  A Long LongSolveLong Longs) the { +memset (bit,0,sizeof(bit)); -     intL=0; $      while(s) { $bit[++l]=s%Ten; -S/=Ten; -     } the     returnDFS (L,0,0,0); - }Wuyi  the intMain () - { Wu     intT; -Cin>>T; Aboutmemset (Prime,0,sizeof(prime)); $ P (); -      while(t--){ -memset (dp,-1,sizeof(DP)); -         Long Longs1,s2; ACin>>s1>>S2; + //cout<<solve (S2) << "" <<solve (s1-1) <<endl; theCout<<solve (S2)-solve (s1-1) <<Endl; -     } $     return 0; the}

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HDU XXXX: Number of primes [l,r] (DP)