[CQOI2016] * （数位dp）

link

$solution:$

$10^{10} leq L leq R < 10^{11}$这个数据范围很容易想到数位$dp$。

依照题意模拟即可。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define int long long
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
int l,r,dig[18],f[12][11][2][5][2][2];
int dfs(int pos,int las,bool done,bool pd,int now,bool pd4,bool pd8){
    //cout<<"pos:"<<pos<<" las:"<<las<<" done:"<<done<<" pd:"<<pd<<" now:"<<now<<endl;
    //printf("pos:%d las:%d done:%d pd:%d now:%d
",pos,las,done,pd,now);
    //system("pause");
    if(pd4==1&&pd8==1) return 0;
    if(pos==0){
        if(pd==1) return 1;
        return 0;
    }
    if(done==0)
        if(f[pos][las][pd][now][pd4][pd8]!=-1) return f[pos][las][pd][now][pd4][pd8];
    int begin,end;
    if(las==11) begin=1;
    else begin=0;
    if(done==1) end=dig[pos];else end=9;
    int ans=0;
    for(int i=begin;i<=end;i++){
        if(las==i){
            if(now>=2) ans+=dfs(pos-1,i,done&&(i==dig[pos]),1,3,pd4|(i==4),pd8|(i==8));
            else ans+=dfs(pos-1,i,done&&(i==dig[pos]),pd,now+1,pd4|(i==4),pd8|(i==8));
        }else ans+=dfs(pos-1,i,done&&(i==dig[pos]),pd,1,pd4|(i==4),pd8|(i==8));
    }
    if(done==0) f[pos][las][pd][now][pd4][pd8]=ans;
    return ans;
}int Dig;
int solve(int x){
    dig[0]=0;
    while(x!=0){
        dig[++dig[0]]=x%10;
        x/=10;
    }
    if(dig[0]!=Dig) return 0;
    return dfs(dig[0],11,1,0,0,0,0);
}
int Qdig(int x){
    dig[0]=0;
    while(x!=0){
       dig[++dig[0]]=x%10;
       x/=10;
    }return dig[0];
}
signed main(){
    memset(f,-1,sizeof(f));
    l=read(),r=read();Dig=Qdig(r);
    printf("%lld
",solve(r)-solve(l-1));
    return 0;
}