HDU 4734 F(x)

F(x)

Problem Description

For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-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



解题思路

给定A，B, 在[0,B]之间的数字假设为i，问你有多少数字 f[i] 值小于f[A]

解题思路：

数位DP，只需按照位数转移为缩短1位的子问题即可

解题代码：

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

const int maxn=50000;
int dp[10][maxn];

int DP(int len,int sum){
    if(sum<0) return 0;
    if(len==0) return 1;
    if(dp[len][sum]!=-1) return dp[len][sum];
    int ans=0;
    for(int i=0;i<=9;i++){
        ans+=DP(len-1,sum-i*(1<<(len-1)) );
    }
    return dp[len][sum]=ans;
}

int f(int x){
    int s=0,sum=0;
    while(x>0){
        sum+=(1<<s)*(x%10);
        x/=10;
        s++;
    }
    return sum;
}

int dfs(int a,int b){
    vector <int> v;
    while(b>0){
        v.push_back(b%10);
        b/=10;
    }
    int ans=0,sum=f(a);
    for(int i=v.size()-1;i>=0;i--){
        for(int t=0;t<v[i];t++){
            ans+=DP(i,sum);
            sum-=(1<<i);
        }
        if(i==0) ans+=DP(i,sum);
    }
    return ans;
}

int main(){
    memset(dp,-1,sizeof(dp));
    int t,a,b;
    cin>>t;
    for(int i=1;i<=t;i++){
        cin>>a>>b;
        cout<<"Case #"<<i<<": "<<dfs(a,b)<<endl;
    }
    return 0;
}