UESTC - 1137 数位DP

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 5e4+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
ll dp[233][2],a[233];
ll dfs(int cur,int limit,int _6){
    if(cur==0) return 1;
    if(~dp[cur][_6]&&!limit) return dp[cur][_6];
    int up=limit?a[cur]:9;
    ll ans=0;
    rep(i,0,up){
        if(i==2&&_6)continue;
        if(i==4)continue;
        ans+=dfs(cur-1,limit&&a[cur]==i,i==6);
    }
    return limit?ans:dp[cur][_6]=ans;
}
ll solve(int x){
    int cur=0;
    while(x){
        a[++cur]=x%10;
        x/=10;
    }
    return dfs(cur,1,0);
}
int main(){
    int l,r;
    memset(dp,-1,sizeof dp);
    while(~scanf("%d%d",&l,&r)){
        if(l==r&&r==0)break;
        println(solve(r)-solve(l-1));
    }
    return 0;
}