HDU 4734F(x) 数位 dp

H - F(x)

Time Limit:500MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4734

Appoint description: System Crawler  (2016-04-28)

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 + A ₁ * 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 < 10 ⁹)

 

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 1001 105 100 

 

Sample Output

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

 

 ACcode:

#include <cstdio>#include <iostream>#include <cstring>using namespace std;int dp[11][10000];int data[11];int cnt=1;int k;int dfs(int len,int sum,int limit){    if(!len)return sum>=0;    if(sum<0)return 0;    if(!limit&&dp[len][sum]!=-1)return dp[len][sum];    int ed=limit?data[len]:9;    int ans=0;    for(int i=0;i<=ed;++i)        ans+=dfs(len-1,sum-i*(1<<(len-1)),limit&&i==ed);    return limit?ans:dp[len][sum]=ans;}void doit(){    int a,b;    scanf("%d%d",&a,&b);    int len=0;    k=0;    while(a){        k+=(a%10)*(1<<len);        a/=10;        len++;    }    len=0;    while(b){        data[++len]=b%10;        b/=10;    }    printf("Case #%d: %d\n",cnt++,dfs(len,k,1));}int main(){    int loop;memset(dp,-1,sizeof(dp));    scanf("%d",&loop);    while(loop--)doit();    return 0;}/*21231 2323 43543 454231 2312100 2332 4212312 1223122 444*/