HDU 4734 F(x)

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5003    Accepted Submission(s): 1864

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

 

Recommend

liuyiding   |   We have carefully selected several similar problems for you:  6014 6013 6012 6011 6010 

 

//初始的是时候sum是f(a),枚举一位就减去这一位在计算f(i)的权值，显然sum=0时就是满足的，后面的位数凑足sum位就可以了
//只有sum==0时，满足条件 
#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e4+5;
int cas,A,b,a[12],dp[12][N],all;
int f(int x){
    if(!x) return 0;
    return (f(x/10)<<1)+(x%10);
}
int dfs(int pos,int sum,bool limit){
    if(!pos) return sum<=all;
    if(sum>all) return 0;
    if(!limit && dp[pos][all-sum]!=-1) return dp[pos][all-sum];
    int up=limit?a[pos]:9;
    int ans=0;
    for(int i=0;i<=up;i++){
        ans+=dfs(pos-1,sum+i*(1<<pos-1),limit && i==a[pos]);
    }
    if(!limit) dp[pos][all-sum]=ans;
    return ans;
}
int solve(int x){
    int pos=0;
    for(;x;x/=10) a[++pos]=x%10;
    return dfs(pos,0,1);
}
int main(){
    memset(dp,-1,sizeof dp);
    scanf("%d",&cas);
    for(int i=1;i<=cas;i++){
        scanf("%d%d",&A,&b);
        all=f(A);
        printf("Case #%d: %d
",i,solve(b)); 
    }
    return 0;
}