百度之星2023初赛R1游记

前言 & 吐槽

非常好比赛，爱来自百度。

先是服务器崩溃导致延期到 16:00 - 19:00。

然后赛时客户端也一直卡，给题目描述翻个页都要卡几秒。

基本是跟榜做的，除了 T8 来不及看之外都看了，但怎么全是一眼题。

此外还有一些槽点。比赛体验总体来说不算很好。

希望百度之星明年还是回去用 bestcoder 吧，这个 matiji 真受不住了。

反正能进决赛了，就不打 R2 R3 了。

T5

签到题。

单组询问复杂度 $O(1)$。

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
voi solve()
{
    uint p,k;scanf("%u%u",&p,&k);
    if(k<p)printf("%u\n",k);
    else if(p==1)puts("1");
    else printf("%u\n",k-k/p+!(k%p));
}
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    uint t;scanf("%u",&t);
    while(t--)solve();
    return 0;
}
 

T1

枚举一下相遇点，跑 bfs 即可。

总复杂度 $O(n)$。

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
ullt a,b,c;
uint n,m,u,v;
std::vector<uint>Way[100005];
uint Dist[3][100005];
voi bfs(uint p,uint*D)
{
    for(uint i=0;i<n;i++)D[i]=-1;
    std::queue<uint>Q;D[p]=0,Q.push(p);
    while(Q.size())
    {
        p=Q.front(),Q.pop();
        for(auto s:Way[p])if(_min(D[s],D[p]+1))
            Q.push(s);
    }
}
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    scanf("%llu%llu%llu%u%u%u%u",&a,&b,&c,&u,&v,&n,&m),c=a+b-c,u--,v--;
    while(m--){uint u,v;scanf("%u%u",&u,&v),Way[--u].push_back(--v),Way[v].push_back(u);}
    bfs(u,Dist[0]),bfs(v,Dist[1]),bfs(n-1,Dist[2]);
    ullt ans=-1llu;
    for(uint i=0;i<n;i++)if(~Dist[0][i]&&~Dist[1][i]&&~Dist[2][i])_min(ans,Dist[0][i]*a+Dist[1][i]*b+Dist[2][i]*c);
    if(~ans)printf("%llu\n",ans);else puts("-1");
    return 0;
}
 

T2

暴力枚举行的切割方案，对列的方案 dp 即可。

复杂度 $O(n^32^n)$。

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
uint S[16][16];
uint Dp[16][16];
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    uint n,k,ans=-1;scanf("%u%u",&n,&k);
    for(uint i=1;i<=n;i++)
    {
        for(uint j=1;j<=n;j++)scanf("%u",S[i]+j),S[i][j]+=S[i][j-1];
        for(uint j=1;j<=n;j++)S[i][j]+=S[i-1][j];
    }
    for(uint i=0;i<(1u<<(n-1));i++)
    {
        std::vector<uint>V;for(uint j=0;j<n-1;j++)if(i>>j&1)V.push_back(j+1);
        if(V.size()>k)continue;
        V.insert(V.begin(),0),V.insert(V.end(),n);
        for(uint i=1;i<=n;i++)
        {
            for(uint j=0;j<=i;j++)Dp[i][j]=-1;
            for(uint j=0;j<i;j++)
            {
                uint s=0;
                for(uint k=1;k<V.size();k++)_max(s,S[i][V[k]]-S[i][V[k-1]]-S[j][V[k]]+S[j][V[k-1]]);
                for(uint k=0;k<=j;k++)_min(Dp[i][k+1],std::max(Dp[j][k],s));
            }
        }
        for(uint i=1;i<=n&&i+V.size()<=k+3;i++)_min(ans,Dp[n][i]);
    }
    printf("%u\n",ans);
    return 0;
}

T3

枚举一下删谁，二分一下答案位置就好了。

预处理一些信息可以做到 $O(n\log n)$。

不知道是题目假了还是数据假了，应为只要 $\ge m$ 即视为结束。

结果罚了两发，还因此在这题花了好久时间。/fn

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
uint T[100005],V[100005];ullt SumM[100005],Sum[100005];
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    uint n,m;scanf("%u%u",&n,&m);if(!m)return puts("0"),0;
    std::vector<std::pair<uint,uint> >P;
    while(n--){uint a,b;scanf("%u%u",&a,&b);if(b)P.push_back({a,b});}
    if(P.size()<=1)return puts("-1"),0;
    std::sort(P.begin(),P.end()),n=P.size();
    for(uint i=0;i<n;i++)T[i]=P[i].first,V[i]=P[i].second,Sum[i+1]=Sum[i]+V[i];
    for(uint i=1;i<n;i++)SumM[i]=SumM[i-1]+Sum[i]*(T[i]-T[i-1]);
    ullt ans=0;
    for(uint i=0;i<n;i++)if(SumM[i]<m)
    {
        uint l=i,r=n-1;
        while(l<r){uint mid=(l+r+1)>>1;if(SumM[mid]-V[i]*(T[mid]-T[i])<m)l=mid;else r=mid-1;}
        _max(ans,T[l]+(m+V[i]*(T[l]-T[i])-SumM[l]+Sum[l+1]-V[i]-1)/(Sum[l+1]-V[i]));
    }
    if(!ans)exit(233);
    printf("%llu\n",ans);
    return 0;
}

T6

$a$ 在 $b$ 前的概率为 $a/(a+b)$，分治 FFT 统计一下贡献即可。

复杂度 $O(n+v\log^2v)$。

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
const ullt Mod=998244353,g=3;
struct modint
{
    ullt v;
    modint():v(0){}
    modint(ullt v):v(v%Mod){}
    ullt&operator()(){return v;}
    friend modint operator+(modint a,modint b){return a()+b();}
    friend modint operator-(modint a,modint b){return a()+Mod-b();}
    friend modint operator*(modint a,modint b){return a()*b();}
    friend modint operator/(modint a,modint b){return a*b.inv();}
    friend modint operator^(modint a,ullt b)
    {
        modint ans(1);
        while(b)
        {
            if(b&1)ans*=a;
            a*=a,b>>=1;
        }
        return ans;
    }
    modint inv(){return(*this)^(Mod-2);}
    modint&operator+=(modint b){return*this=*this+b;}
    modint&operator-=(modint b){return*this=*this-b;}
    modint&operator*=(modint b){return*this=*this*b;}
    modint&operator/=(modint b){return*this=*this/b;}
    modint&operator^=(ullt b){return*this=*this^b;}
    voi read(){scanf("%llu",&v),v%=Mod;}
    voi print(){printf("%llu",v);}
    voi println(){print(),putchar('\n');}
};
typedef std::vector<modint>modvec;
struct NTT
{
    uint n;std::vector<uint>Rev;modvec G;
    voi bzr(uint len)
    {
        n=1;while(n<len)n<<=1;
        Rev.resize(n),G.resize(n);
        modint w=modint(g)^((Mod-1)/n);
        for(uint i=1;i<n;i++)Rev[i]=(Rev[i>>1]|(i&1?n:0))>>1;
        G[0]=1;for(uint i=1;i<n;i++)G[i]=G[i-1]*w;
    }
    voi ntt(modvec&x,bol op)
    {
        if(x.size()<n)x.resize(n);
        for(uint i=1;i<n;i++)if(i<Rev[i])std::swap(x[i],x[Rev[i]]);
        for(uint i=1;i<n;i<<=1)for(uint j=0;j<n;j+=i<<1)for(uint k=0;k<i;k++)
        {
            modint u=x[j|k],v=x[i|j|k]*G[n/i/2*k];
            x[j|k]=u+v,x[i|j|k]=u-v;
        }
        if(op)
        {
            std::reverse(x.begin()+1,x.begin()+n);
            modint v=modint(n).inv();
            for(uint i=0;i<n;i++)x[i]*=v;
        }
    }
};
uint Cnt[100005];
modint S[200005];
voi dfs(uint l,uint r)
{
    if(l==r)return;
    uint mid=(l+r)>>1;
    dfs(l,mid),dfs(mid+1,r);
    modvec A,B;
    for(uint i=l;i<=mid;i++)A.push_back(Cnt[i]);
    for(uint i=mid+1;i<=r;i++)B.push_back(modint(i)*Cnt[i]);
    NTT s;s.bzr(r-l);
    s.ntt(A,0),s.ntt(B,0);
    for(uint i=0;i<A.size();i++)A[i]*=B[i];
    s.ntt(A,1);
    for(uint i=0;i<r-l;i++)S[l+mid+1+i]+=A[i];
}
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    uint n;scanf("%u",&n);
    for(uint i=0,v;i<n;i++)scanf("%u",&v),Cnt[v]++;
    dfs(1,100000);
    modint ans;
    for(uint i=1;i<200000;i++)
        ans+=S[i]/i;
    ans.println();
    return 0;
}

T7

首先肯定是 $k$ 个一位数加一个 $n-k$ 位数。

对于 $n-k$ 很小的情况，可以直接暴力解决。

否则，由于没有 $0$，那个 $n-k$ 位数即为高 $n-k-8$ 位最大的数，使用哈希或者 SA 求出所有合法位置枚举一下即可。

总复杂度 $O(n\log n)$。SA 写挂了好几次。/fn

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
chr C[1000005];uint Rank[1000005],SA[1000005],H[1000005],n;
voi build()
{
    for(uint i=0;i<n;i++)SA[i]=i;
    std::sort(SA,SA+n,[&](uint a,uint b){return C[a]<C[b];});
    for(uint i=n-1;~i;i--)Rank[SA[i]]=i!=n-1&&C[SA[i]]==C[SA[i+1]]?Rank[SA[i+1]]:i;
    for(uint h=1;h<n;h<<=1)
    {
        static uint Cnt[1000005];
        for(uint i=0;i<h;i++)H[i]=n-h+i;
        for(uint i=0,j=h;i<n;i++)if(SA[i]>=h)H[j++]=SA[i]-h;
        for(uint i=0;i<n;i++)Cnt[i]=i;
        for(uint i=n-1;~i;i--)SA[Cnt[Rank[H[i]]]--]=H[i];
        bol ok=true;
        for(uint i=n-1;~i;i--)H[SA[i]]=i!=n-1&&Rank[SA[i]]==Rank[SA[i+1]]
                    &&(SA[i]+h<n?Rank[SA[i]+h]:-1u)==(SA[i+1]+h<n?Rank[SA[i+1]+h]:-1u)?(ok=false,H[SA[i+1]]):i;
        for(uint i=0;i<n;i++)Rank[i]=H[i];
        if(ok)break;
    }
    for(uint i=0,j=0;i<n;i++)
    {
        if(j)j--;
        if(Rank[i])while(i+j<n&&SA[Rank[i]-1]+j<n&&C[i+j]==C[SA[Rank[i]-1]+j])j++;
        H[Rank[i]]=j;
    }
    // for(uint i=0;i<n;i++)puts(C+SA[i]);
}
uint W[1000005];
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    uint k;scanf("%u%u%s",&n,&k,C);
    if(!k)return puts(C),0;
    if(n-k<=8)
    {
        uint ans=0,s=0;for(uint i=0;i<n;i++)s+=C[i]-'0';
        for(uint l=0;l<=k;l++)
        {
            uint t=0,q=0;for(uint j=0;j<n-k;j++)t=t*10+(C[l+j]-'0'),q+=C[l+j]-'0';
            _max(ans,t+s-q);
        }
        printf("%u\n",ans);
        return 0;
    }
    static uint S[1000005];
    for(uint i=0;i<n;i++)S[i+1]=S[i]+(C[i]-'0');
    build();
    uint r=n-1;while(SA[r]>k)r--;
    for(uint i=0;i<n-k;i++)W[i]=C[SA[r]+n-k-i-1]-'0';
    uint w=S[SA[r]+n-k]-S[SA[r]];
    while(H[r]>=n-k-8)if(SA[--r]<=k)_min(w,S[SA[r]+n-k]-S[SA[r]]);
    W[0]+=S[n]-w;
    uint t=0;
    while(W[t])W[t+1]+=W[t]/10,W[t]%=10,t++;
    while(t--)putchar(W[t]+'0');
    putchar('\n');
    return 0;
}

T4

随便拿个平衡树维护一下就好了。复杂度 $O(n\log n)$。

比赛最后半小时时开这题，写完刚好结束了，所以不知道代码有没有写挂。/fn

#include <bits/stdc++.h>
typedef unsigned uint;
typedef long long llt;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _min(T&a,T b){return b<a?(a=b,true):false;}
template<typename T>bol _max(T&a,T b){return a<b?(a=b,true):false;}
// And in that light, I find deliverance.
namespace Splay
{
    struct node
    {
        node*son[2],*fath;uint siz,maxv;std::vector<uint>v;
        node(uint v):son({NULL,NULL}),fath(NULL),siz(1),maxv(v),v({v}){}
        bol howson(){return this==fath->son[1];}
        voi pushup()
        {
            maxv=v.back(),siz=1;
            if(son[0])_max(maxv,son[0]->maxv),siz+=son[0]->siz;
            if(son[1])_max(maxv,son[1]->maxv),siz+=son[1]->siz;
        }
        voi rotate()
        {
            if(!fath)return;
            node*f=fath,*ff=fath->fath;bol sk=howson();
            if((fath=ff))ff->son[f->howson()]=this;
            if((f->son[sk]=son[!sk]))son[!sk]->fath=f;
            (son[!sk]=f)->fath=this,f->pushup(),pushup();
        }
        voi println()
        {
            printf("%u",(uint)v.size());
            for(auto s:v)printf(" %u",s);
            puts("");
        }
    }*rot;
    voi splay(node*p)
    {
        while(p->fath)
        {
            if(p->fath->fath)(p->fath->howson()==p->howson()?p->fath:p)->rotate();
            p->rotate();
        }
        (rot=p)->pushup();
    }
    voi kth(uint k)
    {
        node*p=rot;
        while(true)
        {
            if(p->son[0])
            {
                if(k<p->son[0]->siz){p=p->son[0];continue;}
                k-=p->son[0]->siz;
            }
            if(!k)return splay(p);
            k--,p=p->son[1];
        }
    }
    voi insert(uint k,uint v)
    {
        if(!rot){rot=new node(v);return;}
        if(k<rot->siz)kth(k);
        if(!k)return(rot->son[0]=new node(v))->fath=rot,splay(rot->son[0]);
        node*p=k==rot->siz?rot:rot->son[0];
        while(p->son[1])p=p->son[1];
        (p->son[1]=new node(v))->fath=p,splay(p);
    }
    typedef std::pair<uint,uint>RET;
    RET doit(uint k,uint v)
    {
        if(!rot)return insert(k,v),RET{0,1};
        kth(k);
        if(rot->maxv<=v)return insert(k,v),RET{k+1,rot->siz};
        if(rot->v.back()>v)return rot->v.push_back(v),rot->pushup(),RET{(k+1)%rot->siz,1};
        node*p=rot->son[rot->son[1]&&rot->son[1]->maxv>v];
        while(true)
        {
            if(p->son[0]&&p->son[0]->maxv>v){p=p->son[0];continue;}
            if(p->v.back()>v)break;
            p=p->son[1];
        }
        p->v.push_back(v),splay(p);
        uint r=rot->son[0]?rot->son[0]->siz:0;
        return{(r+1)%rot->siz,r+rot->siz*(r<k)+1-k};
    }
    voi dfs(node*p)
    {
        if(!p)return;
        dfs(p->son[0]);
        p->println();
        dfs(p->son[1]);
    }
    voi print(uint k)
    {
        printf("%u\n",rot->siz);
        kth(k);
        rot->println();
        dfs(rot->son[1]);
        dfs(rot->son[0]);
    }
}
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    freopen("QAQ.out","w",stdout);
#endif
    uint n,k=0;ullt s=0;
    scanf("%u",&n);
    if(!n)return puts("0 0");
    while(n--)
    {
        uint v;scanf("%u",&v);
        auto t=Splay::doit(k,v);
        k=t.first,s+=t.second;
    }
    printf("%llu ",s);
    Splay::print(k);
    return 0;
}