HDU 4734 F (x)

F (x) Time limit:500msmemory limit:32768kbthis problem'll be is judged onHDU. Original id:4734
64-bit integer IO format: %i64d Java class name: Main For a decimal number x with n digits (anan-1an-2 ... A2A1), we define its weight as F (x) = an * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1. Now is given numbers A and B, please calculate how many numbers is there between 0 and B, inclusive, whose weigh T is no more than F (A).InputThe first line has a number T (T <= 10000), indicating the number of test cases.
For each test case, there is numbers A and B (0 <= B < 109)OutputFor every case,you should output ' case #t: ' At first, without quotes. The TIs the case number starting from 1. Then output the answer.Sample Input

30 1001 105 100

Sample Output

Case #1:1Case #2:2Case #3:13

SourceACM/ICPC Asia Regional Chengdu Online Problem solving: Digital DP

1#include <bits/stdc++.h>2 using namespacestd;3 intdp[Ten][200010],pos[Ten];4 intFintx) {5     intb =1, ret =0;6      while(x) {7RET + = XTen*b;8X/=Ten;9b <<=1;Ten     } One     returnret; A } - intDfsintLenintNowBOOLflag) { -     if(Len <0)returnNow >=0; the     if(Now <0)return 0; -     if(!flag && Dp[len][now]! =-1)returnDp[len][now]; -     intD = Flag?pos[len]:9, ret =0; -      for(inti =0; I <= D; ++i) +RET + = DFS (len-1, now-i* (1<<len), flag&&i==d); -     if(!flag) Dp[len][now] =ret; +     returnret; A } at intCalcintAintb) { -     intCNT =0; -      while(b) { -pos[cnt++] = b%Ten; -b/=Ten; -     } in     returnDFS (cnt-1, F (a),true); - } to intMain () { +     intkase,a,b; -scanf"%d",&Kase); thememset (dp,-1,sizeofDP); *      for(intCS =1; CS <= Kase; ++CS) { $scanf"%d%d",&a,&b);Panax Notoginsengprintf"Case #%d:%d\n", Cs,calc (A, b)); -     } the     return 0; +}

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HDU 4734 F (x)