HDU4734 F(x) （数位DP）

F(x)

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

Total Submission(s): 3900 Accepted Submission(s): 1442

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

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#include <cstring>
#include <cstdio>
#include <string.h>
#include <iostream>
#include <cmath>
const int maxn = (1<<10)*10+5;
using namespace std;
int numa[20],numb[20];
int weight;
int dp[20][maxn];

int dfs(int pos,int sum,int over)
{
if(pos<0)
{
return sum>=0;
}
if(sum<0)
return 0;
if(dp[pos][sum]!=-1&&!over)
return dp[pos][sum];
int last=over?numb[pos]:9;
int ans=0;
for(int i=0;i<=last;i++)
{
int now;
now=i*pow(2,pos);
//if(sum<=weight)
ans+=dfs(pos-1,sum-now, over&&i==last);
}
if(!over)
dp[pos][sum]=ans;
return ans;
}
int solve(int a,int b)
{
weight=0;
int lena=0,lenb=0;
while(a)
{
numa[lena++]=a%10;
a/=10;
}
for(int i=lena;i>=1;i--)
{
weight+=numa[i-1]*pow(2,i-1);
}

while(b)
{
numb[lenb++]=b%10;
b/=10;
}
return dfs(lenb-1,weight,true);
}

int main()
{
int cas;
scanf("%d",&cas);
int t=1;
memset(dp,-1,sizeof(dp));
while(cas--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("Case #%d: %d\n",t++,solve(a,b));
}
return 0;
}