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
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define il inline 5 #define Rg register 6 #define go(i,a,b) for(i=a;i<=b;i++) 7 #define yes(i,a,b) for(i=a;i>=b;i++) 8 #define ll long long 9 #define mem(a,b) memset(a,-1,sizeof(a)); 10 using namespace std; 11 il ll read() 12 { 13 ll x=0,y=1;char c=getchar(); 14 while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();} 15 while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();} 16 return x*y; 17 } 18 int len,a[20]; 19 ll l,r,mod,f[20][200][200]; 20 il ll dfs(int pos,int sum,int ys,int limit) 21 { 22 if(pos>len)return (ys==0 && sum==mod); 23 if(!limit && f[pos][sum][ys]!=-1)return f[pos][sum][ys]; 24 ll as=0;int maxs=limit?a[len-pos+1]:9,i; 25 go(i,0,maxs)as+=dfs(pos+1,sum+i,(1LL*ys*10+i)%mod,i==maxs&&limit); 26 if(!limit)f[pos][sum][ys]=as;return as; 27 } 28 il ll sol(ll x) 29 { 30 len=0; 31 while(x)a[++len]=x%10,x/=10; 32 ll as=0; 33 go(mod,1,9*len){mem(f,0);as+=dfs(1,0,0,1);} 34 return as; 35 } 36 int main() 37 { 38 l=read(),r=read(); 39 printf("%lld ",sol(r)-sol(l-1)); 40 return 0; 41 }