51nod 1232 完美数

题目思路：数位dp，若这个数能被每位的非0数整除，那么这个数一定可以被每一位数的lcm整除，lcm（1,2,3,4,5,6,7,8,9） = 2520，所以可以通过将这个数对2520取模来压缩空间,取模结果计做mod

dp[pos][lcm][mod]，显然20*2520*2520仍然过大，所以我们对mod进行离散，再次压缩空间。

#include <iostream>
#include <cmath>
#include <string.h>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
#define LL long long
#define MAXSIZE 2600
#define INF 0x3f3f3f3f

LL dp[][][MAXSIZE];
const LL Mod = ;

int bit[MAXSIZE],ha[MAXSIZE];

LL gcd(LL n,LL m)
{
    if(n%m == )
        return m;
    return gcd(m,n%m);
}

void Init()
{
    int pos = ;
    for(int i=;i<=Mod;i++)
    {
        if(%i == )
            ha[i] = pos++;
    }
}

LL dfs(int pos,int lcm,int mod,int limit)
{
    if(pos < )
    {
        if(mod%lcm == )
            return ;
        return ;
    }

    if(dp[pos][ha[lcm]][mod] != - && !limit)
        return dp[pos][ha[lcm]][mod];

    LL ans = ;
    int len = limit?bit[pos]:;
    for(int i=;i<=len;i++)
    {
        if(i != )
        {
            int new_lcm = lcm*i/(gcd(lcm,i));
            int num = (mod*+i)%Mod;
            ans += dfs(pos-,new_lcm,num,limit&&i==len);
        }

        else
        {
            int num = mod*%Mod;
            ans += dfs(pos-,lcm,num,limit&&i==len);
        }
    }
    if(!limit)
        dp[pos][ha[lcm]][mod] = ans;
    return ans;
}

LL Solve(LL n)
{
    int pos = ;
    while(n)
    {
        bit[pos++] = n%;
        n /= ;
    }
    LL ans = dfs(pos-,,,);
    return ans;
}

int main()
{
    Init();
    int T;
    LL n,m;
    scanf("%d",&T);
    memset(dp,-,sizeof(dp));
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        LL ans = Solve(m) - Solve(n-);
        printf("%lld\n",ans);
    }
    return ;
}