HDU 4734 F(x) 数位 dp

H - F(x)

Time Limit:500MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 4734

Appoint description:  System Crawler  (2016-04-28)

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

      
      
      
      

       
       
       
       3
0 100
1 10
5 100 
      
      
      
      

 

Sample Output

      
      
      
      

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

 

 ACcode:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int dp[11][10000];
int data[11];
int cnt=1;
int k;
int dfs(int len,int sum,int limit){
    if(!len)return sum>=0;
    if(sum<0)return 0;
    if(!limit&&dp[len][sum]!=-1)return dp[len][sum];
    int ed=limit?data[len]:9;
    int ans=0;
    for(int i=0;i<=ed;++i)
        ans+=dfs(len-1,sum-i*(1<<(len-1)),limit&&i==ed);
    return limit?ans:dp[len][sum]=ans;
}
void doit(){
    int a,b;
    scanf("%d%d",&a,&b);
    int len=0;
    k=0;
    while(a){
        k+=(a%10)*(1<<len);
        a/=10;
        len++;
    }
    len=0;
    while(b){
        data[++len]=b%10;
        b/=10;
    }
    printf("Case #%d: %d\n",cnt++,dfs(len,k,1));
}
int main(){
    int loop;memset(dp,-1,sizeof(dp));
    scanf("%d",&loop);
    while(loop--)doit();
    return 0;
}
/*
2123
1 23
23 435
43 454
231 2312
100 23
32 421
2312 1223
122 444
*/