HDOJ 4734 F(x)

数位DP。。。。

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 750    Accepted Submission(s): 286

Problem Description

For a decimal number x with n digits (A _nA _n-1A _n-2 ... A ₂A ₁), we define its weight as F(x) = A _n * 2 ^n-1 + A _n-1 * 2 ^n-2 + ... + A ₂ * 2 + A ₁ * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).

 

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
For each test case, there are two numbers A and B (0 <= A,B < 10 ⁹)

 

Output

For every case,you should output "Case #t: " at first, without quotes. The  t is the case number starting from 1. Then output the answer.

 

Sample Input

30 100
1 10
5 100

 

Sample Output

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

 

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 

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liuyiding

 

 

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; typedef  long  long  int LL; LL dp[ 12][ 111111]; int bit[ 12]; int dfs( int pos, int sum, bool limit) {      if(pos==- 1)  return  1;      if(~dp[pos][sum]&&limit== false)  return dp[pos][sum];      int end=limit?bit[pos]: 9;      int res= 0;      for( int i= 0;i<=end;i++)     {          if((sum-i*( 1<<pos))>= 0)             res+=dfs(pos- 1,sum-i*( 1<<pos),limit&&i==end);     }      if(!limit)         dp[pos][sum]=res;      return res; } int getsum( int x) {      int l= 1,sum= 0;      while(x)     {         sum+=l*(x% 10);         x/= 10; l=l* 2;     }      return sum; } LL colu( int x, int y) {      int pos= 0,sum=getsum(y);      while(x)     {         bit[pos++]=x% 10;         x/= 10;     }      return dfs(pos- 1,sum, true); } int main() {      int cas= 1,x,y,t;     memset(dp,- 1, sizeof(dp));     scanf( "%d",&t);      while(t--)     {         scanf( "%d%d",&y,&x);         printf( "Case #%d: %I64d\n",cas++,colu(x,y));     }      return  0; }

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