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。当中F为斐波那契数列。F1=F2=1。对每一个询问输出答案模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;
}

然而出题人太奇妙，这样的做法常数极大，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]);
}