$BZOJ1799 Luogu4127$ 月之谜 数位统计$DP$

AcWing

Description

Sol

看了很久也没有完全理解直接$DP$的做法,然后发现了记搜的做法,觉得好棒!

这里是超棒的数位$DP$的记搜做法总结   看完仿佛就觉得自己入门了,但是就像文中写的,还是要多做题才能真正弄明白鸭,还要加油(ง •_•)ง

然后这里是同一个人的本题题解

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#define il inline
#define Rg register
#define go(i,a,b) for(i=a;i<=b;i++)
#define yes(i,a,b) for(i=a;i>=b;i++)
#define ll long long
#define mem(a,b) memset(a,-1,sizeof(a));
using namespace std;
il ll read()
{
    ll x=0,y=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*y;
}
int len,a[20];
ll l,r,mod,f[20][200][200];
il ll dfs(int pos,int sum,int ys,int limit)
{
    if(pos>len)return (ys==0 && sum==mod);
    if(!limit && f[pos][sum][ys]!=-1)return f[pos][sum][ys];
    ll as=0;int maxs=limit?a[len-pos+1]:9,i;
    go(i,0,maxs)as+=dfs(pos+1,sum+i,(1LL*ys*10+i)%mod,i==maxs&&limit);
    if(!limit)f[pos][sum][ys]=as;return as;
}
il ll sol(ll x)
{
    len=0;
    while(x)a[++len]=x%10,x/=10;
    ll as=0;
    go(mod,1,9*len){mem(f,0);as+=dfs(1,0,0,1);}
    return as;
}
int main()
{
    l=read(),r=read();
    printf("%lld
",sol(r)-sol(l-1));
    return 0;
}