[莫队算法 线段树 斐波那契 暴力] CF 633H Fibonacci-ish II

题目大意：给出一个长度为n的数列a。对于一个询问lj和rj，将a[lj]到a[rj]从小到大排序后并去重，设得到的新数列为b，长度为k，求F1*b1+F2*b2+F3*b3+...+Fk*bk，其中F为斐波那契数列，F1=F2=1。对每个询问输出答案模m。

区间查询离线 用莫队算法

开棵权值线段树，然后用斐波那契的性质update

F(n+m)=F(n+1)*F(m)+F(n)*F(m-1);

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

int P;
int sx[30005];
int icnt;

inline int Bin(int x){
	return lower_bound(sx+1,sx+icnt+1,x)-sx;
}

struct SEGTREE{
	struct node{
		int k;
		int fk,fk_1;
		int a1,a2;
		friend node operator + (node &A,node &B){
			if (!A.k) return B;
			if (!B.k) return A;
			node ret;
			ret.k=A.k+B.k;
			(ret.fk=(A.fk+A.fk_1)*B.fk+A.fk*B.fk_1)%=P;
			(ret.fk_1=A.fk*B.fk+A.fk_1*B.fk_1)%=P;
			ret.a1=A.a1;
			(ret.a1+=A.fk*B.a2+A.fk_1*B.a1)%=P;
			ret.a2=A.a2;
			(ret.a2+=(A.fk+A.fk_1)*B.a2+A.fk*B.a1)%=P;
			return ret;
		}
	};
	node T[120005];
	int cnt[120005];
	int M,TH;
	inline void Build(int n){
		for (M=1,TH=0;M<n+2;M<<=1,TH++);
	}
	inline int Query(){
		return T[1].a1;
	}
	inline void Change(int s,int r){
		s+=M;
		if (r==1)
		{
			cnt[s]++;
			if (cnt[s]==1)
			{
				T[s].k=1;
				T[s].fk=1;
				T[s].fk_1=0;
				(T[s].a1=sx[s-M])%=P;
				(T[s].a2=sx[s-M])%=P;
				while (s>>=1)
					T[s]=T[s<<1]+T[s<<1|1];
			}
		}
		else if (r==-1)
		{
			cnt[s]--;
			if (cnt[s]==0)
			{
				T[s].k=0;
				T[s].fk=0;
				T[s].fk_1=0;
				T[s].a1=0;
				T[s].a2=0;
				while (s>>=1)
					T[s]=T[s<<1]+T[s<<1|1];
			}
		}
	}
}SEG;

int n,Q,B;
int a[30005],ans[30005];

struct event{
	int x,y,lpos;
	int idx;
	bool operator < (const event &B) const{
		return lpos==B.lpos?y<B.y:lpos<B.lpos;
	}
}eve[30005];

inline void Mos()
{
	int l=1,r=0;
	for (int i=1;i<=Q;i++)
	{
		while (r<eve[i].y) SEG.Change(Bin(a[++r]),1);
		while (r>eve[i].y) SEG.Change(Bin(a[r--]),-1);
		while (l<eve[i].x) SEG.Change(Bin(a[l++]),-1);
		while (l>eve[i].x) SEG.Change(Bin(a[--l]),1);
		ans[eve[i].idx]=SEG.Query();
	}
}

int main()
{
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(n); read(P); B=sqrt(n);
	for (int i=1;i<=n;i++)
		read(a[i]),sx[++icnt]=a[i];
	sort(sx+1,sx+icnt+1);
	icnt=unique(sx+1,sx+icnt+1)-sx-1;
	SEG.Build(icnt);
	read(Q);
	for (int i=1;i<=Q;i++)
	{
		read(eve[i].x); read(eve[i].y);
		eve[i].lpos=(eve[i].x-1)/B+1; eve[i].idx=i;
	}
	sort(eve+1,eve+Q+1);
	Mos();
	for (int i=1;i<=Q;i++)
		printf("%d\n",ans[i]);
	return 0;
}

然而出题人太神奇，这种做法常数极大，还是暴力短小精悍

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e4+5;
pair<int,int> a[maxn];
int ans[maxn],step[maxn],f[maxn],l[maxn],r[maxn],last[maxn];

int main()
{
	freopen("t.in","r",stdin);
	freopen("t1.out","w",stdout);
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i].first),a[i].second=i;
    sort(a+1,a+1+n);
    f[0]=1,f[1]=1;
    for(int i=2;i<=n;i++)
        f[i]=(f[i-1]+f[i-2])%m;
    int q;scanf("%d",&q);
    for(int i=1;i<=q;i++)
    {
        scanf("%d%d",&l[i],&r[i]);
        last[i]=-1;
    }
    for(int i=1;i<=n;i++)
    {
        int d = a[i].first % m;
        for(int j=1;j<=q;j++)
        {
            if(a[i].second<l[j]||a[i].second>r[j])continue;
            if(a[i].first==last[j])continue;
            ans[j]=(ans[j]+f[step[j]++]*d)%m;
            last[j]=a[i].first;
        }
    }
    for(int i=1;i<=q;i++)
        printf("%d\n",ans[i]);
}