温馨提示：本题十分卡常数，我手动开O2才过的。而数据范围不伦不类的n<=30000，常数小的O(n²)居然比O(n√nlogn)跑得快……

考虑插进去一个元素对答案产生的影响。原本数列为Σa[i]f[i]，其中1<=i<=n，然后考虑在k位置插入a[0]，答案显然是a[1]f[1]+a[2]f[2]+...+a[0]f[k]+a[k]f[k+1]+...+a[n]f[n+1]，然后直接区间加斐波那契数显然是不可能的。这时候要向后转移斐波那契数列，(a,b)->(a+b,a)->(2a+b,a+b)->(3a+2b,2a+b)->……系数始终是斐波那契数。

删除时向前转移？系数可以和斐波那契数一起直接求。

我是开O2、O3、Ofast优化才过的，如果能正常过可能会换成常数较小的code吧

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<bits/stdc++.h>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
typedef pair<int,int>pii;
const int N=30086;
struct node{int l,r,id;}q[N];
int n,m,B,Q,mod,f[N],g[N],vis[N],a[N],w[N],sz[N<<2],lazy[N<<2],c[N<<2][2],ans[N];
pii p[N];
bool cmp(node a,node b){return (a.l-1)/B==(b.l-1)/B?a.r<b.r:a.l<b.l;}
void pushup(int rt)
{
    sz[rt]=sz[rt<<1]+sz[rt<<1|1];
    c[rt][0]=(c[rt<<1][0]+c[rt<<1|1][0])%mod;
    c[rt][1]=(c[rt<<1][1]+c[rt<<1|1][1])%mod;
}
void add(int rt,int k)
{
    lazy[rt]+=k;
    int a=c[rt][0],b=c[rt][1];
    if(k>0)c[rt][0]=(1ll*a*f[k+1]+1ll*b*f[k])%mod,c[rt][1]=(1ll*a*f[k]+1ll*b*f[k-1])%mod;
    else k=-k,c[rt][0]=(1ll*a*g[k-1]+1ll*b*g[k])%mod,c[rt][1]=(1ll*a*g[k]+1ll*b*g[k+1])%mod;
}
void pushdown(int rt){if(lazy[rt])add(rt<<1,lazy[rt]),add(rt<<1|1,lazy[rt]),lazy[rt]=0;}
void update(int k,int v,int d,int l,int r,int rt)
{
    if(l==r)
    {
        if(v==-1)c[rt][0]=c[rt][1]=sz[rt]=0;
        else c[rt][0]=1ll*f[v]*a[l]%mod,c[rt][1]=1ll*f[v-1]*a[l]%mod,sz[rt]=1;
        return;
    }
    pushdown(rt);
    int mid=l+r>>1;
    if(k<=mid)update(k,v,d,lson),add(rt<<1|1,d);
    else update(k,v+(v!=-1)*sz[rt<<1],d,rson);
    pushup(rt);
}
void add(int x)
{
    if(!x)return;
    if(!vis[x])update(x,1,1,1,m,1);
    vis[x]++;
}
void del(int x)
{
    if(!x)return;
    vis[x]--;
    if(!vis[x])update(x,-1,-1,1,m,1);
}
int main()
{
    scanf("%d%d",&n,&mod),B=170;
    f[1]=f[2]=1;
    for(int i=3;i<=n+1;i++)f[i]=(f[i-1]+f[i-2])%mod;
    g[1]=1,g[2]=mod-1;
    for(int i=3;i<=n+1;i++)g[i]=(g[i-2]-g[i-1]+mod)%mod;
    for(int i=1,x;i<=n;i++)scanf("%d",&x),p[i]=pii(x,i);
    sort(p+1,p+n+1);
    for(int i=1,lst=1e9+7;i<=n;i++)
    {
        if(lst!=p[i].first)lst=p[i].first,a[++m]=lst%mod;
        w[p[i].second]=m;
    }
    scanf("%d",&Q);
    if(mod==1){while(Q--)puts("0");return 0;}
    for(int i=1;i<=Q;i++)scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i;
    sort(q+1,q+Q+1,cmp);
    for(int i=1,l=0,r=0;i<=Q;i++)
    {
        while(r<q[i].r)add(w[++r]);
        while(r>q[i].r)del(w[r--]);
        while(l<q[i].l)del(w[l++]);
        while(l>q[i].l)add(w[--l]);
        ans[q[i].id]=c[1][0];
    }
    for(int i=1;i<=Q;i++)printf("%d\n",ans[i]);
}

View Code