hdu 4734 【数位DP】

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6391    Accepted Submission(s): 2461

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

3 0 100 1 10 5 100

 

Sample Output

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

 

Source

题意理解：要求在不大于B的区间内找到不大于 F(A)的个数有多少个。

解题思路：这道题是一个数位DP的题，关键在于找到状态转移的条件。我们用dp[pos][sum]来保存结果，pos代表的是当前的位置，sum表示当前的依照F()的公式的和是多少，dp[pos][sum]即表示：长度为pos且权值不大于sum的数的个数。这样就可以让每个数所求的个数，是每个数的固有性质，不再需要每次查询都去memset(dp,-1,sizeof(dp)),如果每次都去memset，那必然超时，只需要一次就够了。

#include
#include
#include
using namespace std;
int a[20];
int dp[20][100010];
int totol;
int dfs(int pos,int sum,int limit)
{
    if(pos==-1)
    {
        if(totol