HDU-4734-F(x)

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

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;
}