hdu4734（数位DP）

时间：2014-08-29 14:43:58 收藏：0 阅读：210

F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1684 Accepted Submission(s): 650

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

Sample Output

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

题意：RT

思路：dp[i][j]表示从最低位到第i位取得的总和 <= j 的方案数，当然在递归的过程中还要判断是否为边界，为边界就不能记忆化了，所以不为边界的时候就可以记录dp值

原文：http://blog.csdn.net/cq_phqg/article/details/38924979