HDU 4734 F[x] (数位dp)

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[20];
int dp[20][6000];
int p[50];
int all;
int dfs(int pos,int sta,bool limit)
{
if(pos==-1)
return 1;
if(!limit&&dp[pos][sta]!=-1)
{
return dp[pos][sta];
}
int up=limit?a[pos]:9;
int sum=0,i;
for(i=0;i<=up;i++)
{
int k=sta-i*p[pos];
if(k<0) continue;
sum+=dfs(pos-1,k,limit&&i==up);
}
if(!limit)
dp[pos][sta]=sum;
return sum;
}
int solve(int x)
{
int cnt=0;
while(x)
{
a[cnt++]=x%10;
x/=10;
}
return dfs(cnt-1,all,1);
}
int main(int argc, char *argv[])
{
int t,i,cs=1;
scanf("%d",&t);
memset(dp,-1,sizeof(dp));
p[0]=1;
for(i=1;i<30;i++)
{
p[i]=p[i-1]*2;
}
while(t--)
{
int A,B;
while(scanf("%d %d",&A,&B)!=EOF)
{
all=0;
int cnt1=0;
while(A)
{
all+=(A%10)*p[cnt1++];
A/=10;
}
printf("Case #%d: ",cs++);
printf("%d\n",solve(B));
}
}
return 0;
}

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6851    Accepted Submission(s): 2649


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