F(x) - 数位dp

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).

InputThe 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)OutputFor 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

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>

using namespace std;
const int N=1e4+5;
int mymax,a[20],dp[12][N];

int f(int x){
    int res=0,tmp,mi=1;
    while(x){
        tmp=x%10;
        res+=tmp*mi;
        mi*=2;
        x/=10;
    }
    return res;
}

int dfs(int pos,int sum,int limit){//sum表示当前加到了多少
    if(pos==-1)return sum>=0;
    if(sum<0)return 0;
    if(!limit&&dp[pos][sum]!=-1)return dp[pos][sum];
    int up=limit?a[pos]:9;
    int ans=0;
    for(int i=0;i<=up;i++){
        ans+=dfs(pos-1,sum-i*(1<<pos),limit&&i==a[pos]);
    }
    if(!limit)dp[pos][sum]=ans;
    return ans;
}

int solve(int x){
    int pos=0;
    while(x){
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,mymax,1);
}
int main(){
    int t,a,b;
    scanf("%d",&t);
    memset(dp,-1,sizeof(dp));
    for(int q=1;q<=t;q++){
        
        scanf("%d%d",&a,&b);
        mymax=f(a);
        printf("Case #%d: %d\n",q,solve(b));
    }
}