数位dp入门(HDU 4734)

 题目给了个f(x)的定义：F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1，Ai是十进制数位，然后给出a,b求区间[0,b]内满足f(i)<=f(a)的i的个数

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
 
using namespace std;
const int N=1e4+5;
int dp[12][N];
int f(int x)//先求出题目所给表达式；
{
    if(x==0) return 0;
    int ans=f(x/10);
    return ans*2+(x%10);
}
int all;
int a[12];
int dfs(int pos,int sum,bool limit)//深搜加数位dp
{
    if(pos==-1) {return sum<=all;}
    if(sum>all) return 0;
    if(!limit && dp[pos][all-sum]!=-1) return dp[pos][all-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][all-sum]=ans;
    return ans;
}
int solve(int x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,0,true);
}
int main()
{
    int a,ri;
    int T_T;
    int kase=1;
    scanf("%d",&T_T);
    memset(dp,-1,sizeof dp);
    while(T_T--)
    {
        scanf("%d%d",&a,&ri);
        all=f(a);
        printf("Case #%d: %d\n",kase++,solve(ri));
    }
    return 0;
}