传送门

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);}elseinsert(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);}elseerase(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);}elseTree[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;
}