HDU 4734 F (x) (Digital DP)

HDU 4734 F (x) (Digital DP)

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 + A1 * 1. now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, aggressive, whose weight is no more than F ().

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 shoshould output "Case # t:" at first, without quotes. TIs the case number starting from 1. Then output the answer.

Sample Input

 30 1001 105 100 

Sample Output

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

Simple Digital DP: first, express A as F (x) And then count the number of digits greater than x.

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           using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair
           
            pil;const int INF = 0x3f3f3f3f;int t,temp;LL l,r;int num[20];LL dp[20][20000];LL dfs(int pos,int sum,int flag){ if(pos==0) return sum>=0; if(sum<0) return 0; if(!flag&&dp[pos][sum]!=-1) return dp[pos][sum]; LL ans=0; int ed=flag?num[pos]:9; for(int i=0;i<=ed;i++) { int s=sum-i*(1<<(pos-1)); ans+=dfs(pos-1,s,flag&&i==ed); } if(!flag) dp[pos][sum]=ans; return ans;}LL solve(LL x){ int pos=0; while(x) { num[++pos]=x%10; x/=10; } return dfs(pos,temp,1);}LL work(LL x){ LL ans=0; int l=0; while(x) { ans+=(x%10)*(1<