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

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

 1 //2016.8.31
 2 #pragma comment(linker, "/STACK:1024000000,1024000000")
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <cstring>
 6 
 7 using namespace std;
 8 
 9 int dp[20][10000], bit[20], a, b;//dp[i][j]表示第i位<=j的数目
10 
11 int F(int a)
12 {
13     int ans = 0, len = 0;
14     while(a)
15     {
16         ans += (a%10)*(1<<len);
17         len++;
18         a /= 10;
19     }
20     return ans;
21 }
22 
23 int dfs(int pos, int num, int fg)
24 {
25     if(pos == -1)return num>=0;
26     if(num < 0)return 0;
27     if(!fg && dp[pos][num]!=-1)//记忆化搜索
28           return dp[pos][num];
29     int ans = 0;
30     int ed = fg?bit[pos]:9;
31     for(int i = 0; i <= ed; i++)
32           ans+=dfs(pos-1, num-i*(1<<pos), fg&&i==ed);
33     if(!fg)dp[pos][num] = ans;
34     return ans;
35 }
36 
37 int solve(int b)
38 {
39     int len = 0;
40     while(b)
41     {
42         bit[len++] = b%10;
43         b /= 10;
44     }
45     int ans = dfs(len-1, F(a), 1);
46     return ans;
47 }
48 
49 int main()
50 {
51     int T, kase = 0;
52     cin>>T;
53     memset(dp, -1, sizeof(dp));
54     while(T--)
55     {
56         scanf("%d%d", &a, &b);
57         int ans = solve(b);
58         printf("Case #%d: %d
", ++kase, ans);
59     }
60 
61     return 0;
62 }