HDOJ 4734 F(x)

数位DP。。。。

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 750    Accepted Submission(s): 286

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

30 100
1 10
5 100

 

Sample Output

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

 

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 

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#include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef long long int LL; LL dp[12][111111]; int bit[12]; int dfs(int pos,int sum,bool limit) {     if(pos==-1) return 1;     if(~dp[pos][sum]&&limit==false) return dp[pos][sum];     int end=limit?bit[pos]:9;     int res=0;     for(int i=0;i<=end;i++)     {         if((sum-i*(1<<pos))>=0)             res+=dfs(pos-1,sum-i*(1<<pos),limit&&i==end);     }     if(!limit)         dp[pos][sum]=res;     return res; } int getsum(int x) {     int l=1,sum=0;     while(x)     {         sum+=l*(x%10);         x/=10; l=l*2;     }     return sum; } LL colu(int x,int y) {     int pos=0,sum=getsum(y);     while(x)     {         bit[pos++]=x%10;         x/=10;     }     return dfs(pos-1,sum,true); } int main() {     int cas=1,x,y,t;     memset(dp,-1,sizeof(dp));     scanf("%d",&t);     while(t--)     {         scanf("%d%d",&y,&x);         printf("Case #%d: %I64d\n",cas++,colu(x,y));     }     return 0; }

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