洛谷

Codeforces

思路

一开始想偏想到了DP，后来发现我SB了……

考虑每个\(a_i<x\)的\(i\)，记录它前一个和后一个到它的距离为\(L_i,R_i\)，那么就有

显然把\(L\)数组翻转一下就是一个FFT了。

最后特判\(k=0\)。

代码

#include<bits/stdc++.h>

clock_t t=clock();

namespace my_std{

    using namespace std;

    #define pii pair<int,int>

    #define fir first

    #define sec second

    #define MP make_pair

    #define rep(i,x,y) for (int i=(x);i<=(y);i++)

    #define drep(i,x,y) for (int i=(x);i>=(y);i--)

    #define go(x) for (int i=head[x];i;i=edge[i].nxt)

    #define templ template<typename T>

    #define sz 800202

    typedef long long ll;

    typedef double db;

    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

    templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}

    templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}

    templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}

    templ inline void read(T& t)

    {

        t=0;char f=0,ch=getchar();double d=0.1;

        while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();

        while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();

        if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}

        t=(f?-t:t);

    }

    template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}

    char sr[1<<21],z[20];int C=-1,Z=0;

    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}

    inline void print(register int x)

    {

    	if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;

    	while(z[++Z]=x%10+48,x/=10);

    	while(sr[++C]=z[Z],--Z);sr[++C]='\n';

    }

    void file()

    {

        #ifndef ONLINE_JUDGE

        freopen("a.in","r",stdin);

        #endif

    }

    inline void chktime()

    {

        #ifndef ONLINE_JUDGE

        cout<<(clock()-t)/1000.0<<'\n';

        #endif

    }

    #ifdef mod

    ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}

    ll inv(ll x){return ksm(x,mod-2);}

    #else

    ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}

    #endif

//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}

}

using namespace my_std;

int n,x;

int a[sz];

int L[sz],R[sz];

const db PI=acos(-1.0);

struct Complex

{

    db x,y;

    Complex(db xx=0,db yy=0){x=xx,y=yy;}

};

const Complex operator + (const Complex &a,const Complex &b) {return Complex(a.x+b.x,a.y+b.y);}

const Complex operator - (const Complex &a,const Complex &b) {return Complex(a.x-b.x,a.y-b.y);}

const Complex operator * (const Complex &a,const Complex &b) {return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}

int limit,r[sz];

void FFT_init(int n)

{

    int l=-1;limit=1;

    while (limit<=n+n) ++l,limit<<=1;

    rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);

}

void FFT(Complex *a,int type)

{

    for (int i=0;i<limit;i++) if (i<r[i]) swap(a[i],a[r[i]]);

    for (int mid=1;mid<limit;mid<<=1)

    {

        Complex Wn(cos(PI/mid),type*sin(PI/mid));

        for (int len=mid<<1,j=0;j<limit;j+=len)

        {

            Complex w(1,0);

            for (int k=0;k<mid;k++,w=w*Wn)

            {

                Complex x=a[j+k],y=w*a[j+mid+k];

                a[j+k]=x+y;a[j+k+mid]=x-y;

            }

        }

    }

}

Complex _a[sz],_b[sz];

ll ans[sz];

ll CC(int n){return 1ll*n*(n+1)/2;}

int main()

{

    file();

    int cnt,c,cc;

    read(n,x);

    rep(i,1,n) read(a[i]);

    cnt=c=0;

    rep(i,1,n)

        if (a[i]<x) L[++c]=cnt+1,cnt=0;

        else ++cnt;

    cnt=0;cc=c;

    drep(i,n,1)

        if (a[i]<x) R[c--]=cnt+1,cnt=0;

        else ++cnt;

    reverse(L+1,L+cc+1);

    rep(i,1,cc) _a[i].x=L[i],_b[i].x=R[i];

    FFT_init(cc+1);

    FFT(_a,1);FFT(_b,1);

    rep(i,0,limit-1) _a[i]=_a[i]*_b[i];

    FFT(_a,-1);

    rep(i,1,n) ans[i]=ll(_a[i+cc].x/limit+0.5);

    if (cc)

    {

        rep(i,1,cc) ans[0]+=CC(R[i]-1);

        ans[0]+=CC(L[cc]-1);

    } else ans[0]=CC(n);

    rep(i,0,n) printf("%lld ",ans[i]);

    return 0;

}