HDU 4734 F (x) (digital DP)

F (x)

Time limit:1000/500 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2190 Accepted Submission (s): 828

Problem descriptionfor 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 INPUT30 1001 105 100

Sample outputcase #1:1Case #2:2Case #3:13

Source2013 ACM/ICPC Asia Regional Chengdu Online

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1#include <cstdio>2#include <algorithm>3#include <iostream>4#include <cmath>5#include <cstring>6 #definell Long Long7 8 using namespacestd;9 Ten intN; One intp[ A]; A intb; - intFA,FB; - intnum[ A]; the intdp[ A][25010]; -  - voidinit () { -p[0]=1; +      for(intI=1; i<=Ten; i++) -p[i]=p[i-1]*2; + } A  at intDfsintIintSBOOLe) { -     if(i==-1)returns>=0; -     if(s<0)return 0; -     if(!e&&dp[i][s]!=-1)returnDp[i][s]; -     intU=e?num[i]:9; -     intres=0; in      for(intD=0; d<=u; d++) { -         intnews=s-d*P[i]; toRes+=dfs (I-1, news,e&&d==u); +     } -     returne?res:dp[i][s]=Res; the } *  $ intSolveintx) {Panax Notoginseng     intlen=0; -      while(x) { thenum[len++]=x%Ten; +X/=Ten; A     } the     returnDFS (len-1, FA,1); + } -  $ intMain () { $     //freopen ("In.txt", "R", stdin); -memset (dp,-1,sizeofDP); - init (); the     intT; -     intCa=1;WuyiCin>>T; the      while(t--) { -scanf"%d%d",&a,&b); Wu         intk=0; -Has0; About          while(a) { $fa+=p[k]* (a%Ten); -A/=Ten, k++; -         } -         //cout<< "fa=" <<fa<<endl; Aprintf"Case #%d:%d\n", ca++, solve (b)); +     } the     return 0; -}

HDU 4734 F (x) (digital DP)