hdu 4734 F(x)（数位dp，记忆化深搜，灵活题）

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

   
   
   
   

    
    
    
    3
0 100
1 10
5 100
   
   
   
   

 

Sample Output

   
   
   
   

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

一个N位的数X (A_nA_n-1A_n-2 ... A₂A₁)，F(x) = A_n * 2^n-1 + A_n-1 * 2^n-2 + ... + A₂ * 2 + A₁ * 1，求当x分别取0到B（包括B），共有多少个数符合F(B)<=F(A)

一开始把memset放在while(t--)里，结果一直超时。以后记住，记忆化深搜的东西是可以在多组数据间通用的。

由F(x)的公式很容易想出该用数位dp，dp[pos][num]表示x从1位取到pos位数，F(x)的值比F(A)小（或等于）的有几个。num=F(A)-当前数x的F(x)。num>=0表示F(x)的值比F(A)小（或等于）。

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

#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<vector>  
#include<cmath>  
#include<queue>  
#include<string.h>  
#include<stdlib.h>  
#include<cstdio>  
#define ll long long  
using namespace std;
ll a,b;
ll dp[11][200000];
int bit[11];
int dfs(int pos,int num,int flag){ //flag是个妙点
    if(num<0)
        return 0;
    else if(pos==-1)
        return 1;
    if(flag==0&&dp[pos][num]!=-1)
        return dp[pos][num];
    ll s=0;
    int end=(flag==1?bit[pos]:9); //flag=0表示这个数字前面的位上已经放了比B的那一位小的数字，因此这个数字肯定比B小，因此后面的位能任意取0-9
    for(int i=0;i<=end;++i)       //遍历这一位上能放的所有数字
        s+=dfs(pos-1,num-i*(1<<pos),flag&&i==end); //i==end是针对flag=1的时候,从前面数到这一位还没有比B小
    if(flag==0)  //注意条件
        dp[pos][num]=s;
    return s;
}
int main(){
    int t;
    scanf("%d",&t);
    int cnt=0;
    memset(dp,-1,sizeof(dp)); //记忆化深搜（结果通用） 
    while(t--){
        int a,b;
        scanf("%d%d",&a,&b);
        int l=-1;
        while(b){
            bit[++l]=b%10; //存B每一位上的数
            b/=10;
        }
        int FA=0,len=-1;
        while(a){
            FA+=(a%10)*(1<<(++len); //算F(A)
            a/=10;
        }
        printf("Case #%d: %d\n",++cnt,dfs(l,FA,1));
    }
    return 0;
}