$BZOJ1799\ Luogu4127$ 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=,y=;char c=getchar();
     while(c<''||c>''){if(c=='-')y=-;c=getchar();}
     while(c>=''&&c<=''){x=(x<<)+(x<<)+c-'';c=getchar();}
     return x*y;
 }
 int len,a[];
 ll l,r,mod,f[][][];
 il ll dfs(int pos,int sum,int ys,int limit)
 {
     if(pos>len)return (ys== && sum==mod);
     if(!limit && f[pos][sum][ys]!=-)return f[pos][sum][ys];
     ll as=;int maxs=limit?a[len-pos+]:,i;
     go(i,,maxs)as+=dfs(pos+,sum+i,(1LL*ys*+i)%mod,i==maxs&&limit);
     if(!limit)f[pos][sum][ys]=as;return as;
 }
 il ll sol(ll x)
 {
     len=;
     while(x)a[++len]=x%,x/=;
     ll as=;
     go(mod,,*len){mem(f,);as+=dfs(,,,);}
     return as;
 }
 int main()
 {
     l=read(),r=read();
     printf("%lld\n",sol(r)-sol(l-));
     return ;
 }