hdu 4734 数位dp

http://acm.hdu.edu.cn/showproblem.php?pid=4734

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 1001 105 100

 

Sample Output

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

/**hdu 4734  数位dp 记忆化搜索题目大意：求给定区间1~B中所有小于f（A）的数。解题思路：记忆化搜索， 标记dp[i][j]表示i位数比j小的数的个数。时间卡的很紧，不用记忆化会超时的*/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;int A,B,dp[11][50000],a[25];int len;int f(int n){    int ans=0,len=1;    while(n)    {        ans+=n%10*len;        len*=2;        n/=10;    }   /// printf("f(a)=%d\n",ans);    return ans;}int dfs(int len,int ans,int flag){    if(len<0) return ans>=0;    if(ans<0) return 0;    int sum=0;    if(!flag&&dp[len][ans]!=-1)return dp[len][ans];    int end=flag?a[len]:9;    for(int i=0; i<=end; i++)    {        sum+=dfs(len-1,ans-i*(1<<len),flag&&i==end);    }    if(!flag)dp[len][ans]=sum;    return sum;}int main(){    int T,tt=0;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    while(T--)    {        scanf("%d%d",&A,&B);        len=0;        while(B)        {            a[len++]=B%10;           /// printf("%d ",a[len-1]);            B/=10;        }        ///printf("\n");        printf("Case #%d: %d\n",++tt,dfs(len-1,f(A),1));    }    return 0;}