这题除了暴力踩标程和正解卡常数以外是道很好的题目
首先看到我们要求的东西与\(Fibonacci\)有关，考虑矩阵乘法进行维护。又看到\(n \leq 30000\)，这告诉我们正解算法其实比较暴力，又因为直接在线解决看起来就比较麻烦，所以考虑离线询问，莫队解决。
我们设斐波那契数列的转移矩阵为\(T = \left( \begin{array}{ccc} 0 & 1 \\ 1 & 1 \end{array} \right)\)
先将\(a\)离散化，用一棵线段树维护矩阵运算。那么我们需要支持的是：插入一个数并使比它大的数对应的\(Fibonacci\)数向后移一个。这个可以在线段树的对应节点打上一个\(T\)的标记，表示向右转移一个，经过这个节点时pushdown下去。删除一个数就打上它的逆矩阵的标记。总复杂度为\(O(n\sqrt{n}logn)\)
Tips：如果你TLE在了第35个点，请尽力卡常，简化取模过程、避免不必要运算（详见代码中pushup过程）
#include<bits/stdc++.h>
//This code is written by Itst
#define lch (x << 1)
#define rch (x << 1 | 1)
#define mid ((l + r) >> 1)
using namespace std;
inline int read(){
int a = 0;
char c = getchar();
bool f = 0;
while(!isdigit(c)){
if(c == '-')
f = 1;
c = getchar();
}
while(isdigit(c)){
a = (a << 3) + (a << 1) + (c ^ '0');
c = getchar();
}
return f ? -a : a;
}
const int MAXN = 3e4 + 7;
int step = 0 , N , M , Q , T , cnt , num[MAXN] , lsh[MAXN] , times[MAXN] , ans[MAXN];
struct query{
int ind , l , r;
bool operator <(const query a)const{
return l / T == a.l / T ? ((l / T) & 1 ? r > a.r : r < a.r) : l < a.l;
}
}now[MAXN];
struct matrix{
int a[2][2];
int* operator [](int x){return a[x];}
matrix(bool f = 1){if(f) memset(a , 0 , sizeof(a));}
matrix operator *(matrix b){
matrix c;
for(int i = 0 ; i < 2 ; ++i)
for(int j = 0 ; j < 2 ; ++j)
for(int k = 0 ; k < 2 ; ++k)
c[i][j] += a[i][k] * b[k][j];
for(int i = 0 ; i < 2 ; ++i)
for(int j = 0 ; j < 2 ; ++j)
c[i][j] %= M;
return c;
}
matrix operator *(int b){
matrix c(0);
for(int i = 0 ; i < 2 ; ++i)
for(int j = 0 ; j < 2 ; ++j)
c[i][j] = a[i][j] * b;
return c;
}
matrix operator +(matrix b){
matrix c(0);
for(int i = 0 ; i < 2 ; ++i)
for(int j = 0 ; j < 2 ; ++j)
c[i][j] = (a[i][j] + b[i][j]) % M;
return c;
}
bool operator ==(matrix b){
for(int i = 0 ; i < 2 ; ++i)
for(int j = 0 ; j < 2 ; ++j)
if(a[i][j] != b[i][j])
return 0;
return 1;
}
bool operator !=(matrix b){
return !(*this == b);
}
}F , E , G , a , b;
struct node{
matrix ans , mark;
int times;
}Tree[MAXN << 2];
inline void mark(int x , const matrix mark){
Tree[x].mark = Tree[x].mark * mark;
Tree[x].ans = Tree[x].ans * mark;
}
inline void pushdown(int x){
if(Tree[x].mark != E){
mark(lch , Tree[x].mark);
mark(rch , Tree[x].mark);
Tree[x].mark = E;
}
}
inline void pushup(int x){
a = Tree[lch].ans;
b = Tree[rch].ans;
if(Tree[lch].times != 1)
a = a * Tree[lch].times;
if(Tree[rch].times != 1)
b = b * Tree[rch].times;
Tree[x].ans = a + b;
}
void insert(int x , int l , int r , int tar){
if(l == r){
Tree[x].times = lsh[tar];
return;
}
pushdown(x);
if(mid >= tar){
insert(lch , l , mid , tar);
mark(rch , F);
}
else
insert(rch , mid + 1 , r , tar);
pushup(x);
}
void erase(int x , int l , int r , int tar){
if(l == r){
Tree[x].times = 0;
return;
}
pushdown(x);
if(mid >= tar){
erase(lch , l , mid , tar);
mark(rch , G);
}
else
erase(rch , mid + 1 , r , tar);
pushup(x);
}
void init(int x , int l , int r){
Tree[x].times = l != r;
Tree[x].mark = E;
if(l != r){
init(lch , l , mid);
init(rch , mid + 1 , r);
}
else
Tree[x].ans = F;
}
inline void add(int a){
if(!times[a]++)
insert(1 , 1 , cnt , a);
++step;
}
inline void del(int a){
if(!--times[a])
erase(1 , 1 , cnt , a);
++step;
}
int main(){
N = read();
M = read();
T = sqrt(N);
E[0][0] = E[1][1] = F[0][1] = F[1][0] = F[1][1] = G[1][0] = G[0][1] = 1;
G[0][0] = M - 1;
for(int i = 1 ; i <= N ; ++i)
num[i] = lsh[i] = read();
sort(lsh + 1 , lsh + N + 1);
cnt = unique(lsh + 1 , lsh + N + 1) - lsh - 1;
for(int i = 1 ; i <= N ; ++i)
num[i] = lower_bound(lsh + 1 , lsh + cnt + 1 , num[i]) - lsh;
for(int i = 1 ; i <= cnt ; ++i)
lsh[i] %= M;
Q = read();
for(int i = 1 ; i <= Q ; ++i){
now[i].ind = i;
now[i].l = read();
now[i].r = read();
}
sort(now + 1 , now + Q + 1);
int L = 1 , R = 0;
init(1 , 1 , cnt);
for(int i = 1 ; i <= Q ; ++i){
while(R < now[i].r)
add(num[++R]);
while(L > now[i].l)
add(num[--L]);
while(R > now[i].r)
del(num[R--]);
while(L < now[i].l)
del(num[L++]);
ans[now[i].ind] = Tree[1].ans[0][1] * Tree[1].times % M;
}
cerr << step << endl;
for(int i = 1 ; i <= Q ; ++i)
printf("%d\n" , ans[i]);
return 0;
}