CF633H Fibonacci-ish II 莫队、线段树、矩阵乘法

标签：cci   char   ini   mat   \n   pac   mes   +=   tree   

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);
    }
    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;
}

CF633H Fibonacci-ish II 莫队、线段树、矩阵乘法

标签：cci   char   ini   mat   \n   pac   mes   +=   tree   

原文地址：https://www.cnblogs.com/Itst/p/10262139.html