HDU4734(数位dp)

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4423    Accepted Submission(s): 1632

Problem Description

For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-2 + ... + A₂ * 2 + A₁ * 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 < 10⁹)

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

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 //2016.8.31
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 int dp[][], bit[], a, b;//dp[i][j]表示第i位<=j的数目

 int F(int a)
 {
     int ans = , len = ;
     while(a)
     {
         ans += (a%)*(<<len);
         len++;
         a /= ;
     }
     return ans;
 }

 int dfs(int pos, int num, int fg)
 {
     if(pos == -)return num>=;
     if(num < )return ;
     if(!fg && dp[pos][num]!=-)//记忆化搜索
           return dp[pos][num];
     int ans = ;
     int ed = fg?bit[pos]:;
     for(int i = ; i <= ed; i++)
           ans+=dfs(pos-, num-i*(<<pos), fg&&i==ed);
     if(!fg)dp[pos][num] = ans;
     return ans;
 }

 int solve(int b)
 {
     int len = ;
     while(b)
     {
         bit[len++] = b%;
         b /= ;
     }
     int ans = dfs(len-, F(a), );
     return ans;
 }

 int main()
 {
     int T, kase = ;
     cin>>T;
     memset(dp, -, sizeof(dp));
     while(T--)
     {
         scanf("%d%d", &a, &b);
         int ans = solve(b);
         printf("Case #%d: %d\n", ++kase, ans);
     }

     return ;
 }