CF-55-D-数位DP

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-23

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55-D 题目大意

给定区间$[l,r]$，问区间中有多少个数$x$满足：对于它的每一个非$0$位上的数$y$，都有$y|x$。

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Solution

经典的数位DP题型。

记录状态：$f[pos][num][lcm]$表示填完前$pos$位上的数，这些数构成了数$num$，各个非$0$位数字的最小公倍数为$lcm$，如果数$lcm|num$，那么$num$也就能整除各个非$0$位上的数。

考虑$num$和$lcm$的上界，$1,2···9$的最小公倍数为$2520$，则在记忆化搜索的过程中$num$不断对$2520$取模即可。另一方面，$2520$的因子数量不超过$50$个，也就是$lcm$的种类不超过$50$个，对这些$lcm$做一次映射即可。

剩下的就是套个数位DP的壳即可，因为存在多组测试数据，对于前$pos$位如果都已经达到了上限的话不用记忆化在$f$中。记忆化的话会导致下组测试数据使用到不合法的状态。

时间复杂度为$O(2520+10·50·2520·logU+TlogU)$，实际根本无法跑满。

#include<bits/stdc++.h>
using namespace std;
using ll=long long;

ll f[20][2600][50];
int id[2600],idx=0;

void solve(){
    ll l,r;
    cin>>l>>r;
    l--;
    string x=to_string(l),y=to_string(r);
    function<ll(int,int,int,int,string&)> dp=[&](int pos,int num,int full,int lcm,string &s)->ll{
        if(pos<0){
            return num%lcm==0;
        }
        if(pos>=s.size()){
            return dp(pos-1,num,full,lcm,s);
        }
        if(!full&&f[pos][num][id[lcm]]!=-1){
            return f[pos][num][id[lcm]];
        }
        ll res=0;
        for(int i=0;i<=9;i++){
            if(full&&i>s[pos]-'0') break;
            int nxtfull=full&&(i==s[pos]-'0');
            int nxtlcm=lcm;
            if(i) nxtlcm=lcm*i/__gcd(lcm,i);
            int nxtnum=(num*10+i)%2520;
            res+=dp(pos-1,nxtnum,nxtfull,nxtlcm,s);
        }
        if(!full){
            f[pos][num][id[lcm]]=res;
        }
        return res;
    };
    reverse(x.begin(),x.end());
    reverse(y.begin(),y.end());
    ll high=dp(19,0,1,1,y);
    ll low=dp(19,0,1,1,x);
    cout<<high-low<<'\n';
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    for(int i=1;i<=2520;i++){
        if(2520%i==0){
            id[i]=idx++;
        }
    }
    memset(f,-1,sizeof f);
    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
    return 0;
}