HDU - 2089 数位DP 初步

中文题目，不要62和4

从高位往低位DP，注意有界标志limit的传递

dp2记忆有界情况下的计数结果，据说用处不大

我所参考的入门文章就是半搜索（有界）半记忆（无界）的

进阶指南中提出dfs维度有多少那么dp维度就是多少，因此原文的pre确实是没用的

#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 = 1e3+11;

const double eps = 1e-10;

typedef long long ll;

const int oo = 0x3f3f3f3f;

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[maxn][2],a[maxn],l,r;

ll dp2[maxn][2];

ll DP(int cur,bool _6,bool limit){

    if(cur==0) return 1;

    if(dp[cur][_6]!=-1&&!limit)return dp[cur][_6];

    if(dp2[cur][_6]!=-1&&limit)return dp2[cur][_6];

    int up=(limit?a[cur]:9);

    ll ans=0;

    rep(i,0,up){

        if(i==4)continue;

        if(i==2&&_6)continue;

        ans+=DP(cur-1,i==6,limit&&a[cur]==i);

    }

    return limit?dp2[cur][_6]=ans:(dp[cur][_6]=ans);

}

ll solve(ll num){

    memset(a,0,sizeof a);

    memset(dp,-1,sizeof dp);

    memset(dp2,-1,sizeof dp2);

    int cur=0;

    while(num){

        a[++cur]=num%10;

        num/=10;

    }

    return DP(cur,0,1);

}

int main(){

    while(cin>>l>>r){

        if(l==0&&r==0)break;

        println(solve(r)-solve(l-1));

    }

    return 0;

}