BZOJ 2081: [Poi2010]Beads

Description

问把n截成每个长度后不同子串个数.

Sol

调和极数＋Hash.

首先这是一个式子 \(n\sum_{i=1}^n \frac {1}{i}\) .

这东西就是调和极数再乘上 \(n\) ,他趋近于\(nlnn\)

正反哈希一下.

Code

/**************************************************************
    Problem: 2081
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:6420 ms
    Memory:58552 kb
****************************************************************/

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

#define debug(a) cout<<#a<<"="<<a<<" "

typedef long long LL;
const int N = 2e5+50;
const LL bs = 666671;
const LL p = 5831801;

LL n,ans,ans1;
LL a[N],b[N],rb[N],v[2][N];
vector< int > as;

struct HashTable {
    int h[p],nxt[p],cnt;
    int q[N],t;
    LL v[N];

    void insert(LL x,LL val) {
        v[++cnt]=val,nxt[cnt]=h[x],h[x]=cnt;
        q[++t]=x;
    }
    LL find(LL x,LL val) {
        for(int i=h[x];i;i=nxt[i]) if(v[i]==val) return 1;
        return -1;
    }
    void clear() {
        for(int i=1;i<=t;i++) h[q[i]]=0;
        cnt=0,t=0;
    }
}hh;

inline LL in(LL x=0,char ch=getchar()) { while(ch>'9' || ch<'0') ch=getchar();
    while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();return x; }
inline LL Pow(LL a,LL b,LL r=1) { for(;b;b>>=1,a=a*a%p) if(b&1) r=r*a%p;return r; }
LL GetVal(int kd,LL l,LL r) {
    if(!kd) return (v[kd][r]-v[kd][l-1]+p)%p*rb[l-1]%p;
    else return (v[kd][l]-v[kd][r+1]+p)%p*rb[n-r]%p;
}
int main() {
    n=in();
    for(int i=1;i<=n;i++) a[i]=in();
    b[0]=b[1]=1;for(int i=2;i<=n;i++) b[i]=b[i-1]*bs%p;
    rb[0]=1,rb[1]=Pow(bs,p-2);for(int i=2;i<=n;i++) rb[i]=rb[i-1]*rb[1]%p;

    for(int i=1;i<=n;i++) v[0][i]=(v[0][i-1]+a[i]*b[i]%p)%p;
    for(int i=n;i>=1;i--) v[1][i]=(v[1][i+1]+a[i]*b[n-i+1]%p)%p;

    for(int i=1;i<=n;i++) {
        int r=0;
        hh.clear();
//      cout<<i<<":"<<endl;
        for(int j=1;j+i-1<=n;j+=i) {
            LL vv1=GetVal(0,j,j+i-1),vv2=GetVal(1,j,j+i-1);
//          debug(vv1),debug(vv2)<<endl;

            if(vv1>vv2) swap(vv1,vv2);
            if(hh.find(vv1,vv2)==-1) hh.insert(vv1,vv2),r++;
        }
//      debug(r)<<endl;
//      cout<<"---------------------"<<endl;
        if(r>ans) as.clear(),as.push_back(i),ans=r;
        else if(r==ans) as.push_back(i);
    }
    ans1=as.size();
    printf("%lld %lld\n",ans,ans1);
    for(int i=0;i<ans1;i++) printf("%d%c",as[i]," \n"[i==ans1-1]);
    return 0;
}

/*
21
1 1 1 2 2 2 3 3 3 1 2 3 3 1 2 2 1 3 3 2 1
*/