HDU-4734-F(x) - c++编程基础

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

Sample Output

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

Source 2013 ACM/ICPC Asia Regional Chengdu Online
一道很简单的数位DP！！！！DP[pos][sum] 记录的是长度为pos小于等于sum的个数，这样就不用每次都初始化，我一开始DP记录的是长度为pos，前len-pos个数和sum的个数，因为每组样例DP的值不一定相同如 1 10 dp[0][0] = 1; 而 2 100 dp[0][0] = 2;因此每次都需要初始化，这也是导致超时的原因。这是一道很好的DP题！

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          using namespace std; int dp[12][4600]; int A,B; int F; vector 
         
           digit; void init(){ F = 0; int t = 0; while(A){ F += (A%10)*(1<
          
            F) break; res += dfs(pos-1,val + i*(1<
           
            > ncase; memset(dp,-1,sizeof dp); while(ncase--){ cin >> A >> B; init(); printf("Case #%d: %d\n",T++,solve(B)); } return 0; }