HDU-4734-F(x)

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
一道很簡單的數位DP！！！！DP[pos][sum] 記錄的是長度為pos小於等於sum的個數，這樣就不用每次都初始化，我一開始DP記錄的是長度為pos，前len-pos個數和sum的個數，因為每組樣例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，這也是導致超時的原因。 這是一道很好的DP題！

#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;
}