P2602 [ZJOI2010] 数字计数

//URL:https://www.luogu.com.cn/problem/P2602
/*
数位dp
1.对于某一位 不考虑前导0: f[i]=f[i-1]*10(rt)+1*10^(i-1)(lft)
记忆化搜索 转化到solve(b)-solve(a-1)
dfs(pos,sum,limit,zero,dig):当前位 次数 有无限制 有无前导0 当前搜索dig
1.f[pos][limit][zero][sum] 可以确定已知 答案
2.zero limit 都有要求 可以省略
3.f[pos][sum] [sum]为了不相互干扰 搜索到答案

*/
/*
1 99
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9 20 20 20 20 20 20 20 20 20

*/

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string.h>
#include<queue>
#include<vector> 
#include<bits/stdc++.h>
typedef long long ll;
#define ddd printf("-----------------------\n");
using namespace std;
const int maxn=1e1 +10;
const int mod=998244353;
const int inf=0x3f3f3f3f;

ll dp[20][20],dim[20];
ll dfs(int pos,int sum,int limit,int zero,int dig)// limit 1/0 zero 1/0
{
    if(pos==0) return sum;
    if(limit==0&&zero==0&&dp[pos][sum]!=-1) return dp[pos][sum];
    int end=limit? dim[pos]:9;
    ll res=0;
    for(int i=0;i<=end;i++)
    {
        res+=dfs(pos-1,sum+((zero==0||i)&&(i==dig)),limit&&i==end,zero&&i==0,dig);
    }
    if(limit==0&&zero==0) dp[pos][sum]=res;
    return res;
}
//数位dp的操作
long long  work(long long x,int dig)
{
    memset(dp,-1,sizeof(dp));//初始化
    dim[0]=0;
    while(x)
    {
        dim[++dim[0]]=x%10;//最高位在第一位 
        x/=10;
    }
    return    dfs(dim[0],0,1,1,dig);
}
int main()
{
    ios::sync_with_stdio(false);
    ll a,b; cin>>a>>b;
    for(int i=0;i<=9;i++)//九个digit
    {
         if(i!=9)cout<<work(b,i)-work(a-1,i)<<" ";
         else cout<<work(b,i)-work(a-1,i);
         //我也不知道最后边输出了空格会不会炸（手动滑稽）
    }
    return 0;
}