CF718C Sasha and Array

Description

洛谷传送门

Solution

转移方程就是斐波那契数列求和，题目里也都给了。

矩阵也比较基础吧，不写了。

但是这道题需要用到线段树维护矩阵乘法。

听着挺吓人的，其实也没有多难。

我们首先建一棵矩阵类型的线段树。

然后 \(build\) 为初始输入的斐波那契数（即 \(f^{A - 1}\)，因为我们的 \(f\) 矩阵中已经有前两项了，所以次方要 -1）。

区间修改（加上 \(x\)）其实就是乘上 \(x\) 次方。\(lazy\) 标记中存初始矩阵 \(x\) 次方之后的结果。

\(pushdown\) 时，令原来的数直接乘上 \(lazy\) 标记即可。

查询操作，我们线段树中维护的就是区间和，直接查询即可。

Code

代码个人感觉还是比较简明易懂的，不写注释了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define ll long long

using namespace std;

inline ll read(){
    ll x = 0;
    char ch = getchar();
    while(ch < '0' || ch > '9') ch = getchar();
    while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x;
}

const ll mod = 1e9 + 7;

struct matrix{
    ll num[5][5];
    matrix(){
        memset(num, 0, sizeof(num));
    }
    void init(){
        for(int i = 1; i <= 2; i++)
            num[i][i] = 1;
    }
    bool empty(){
        if(num[1][1] != 1 || num[2][2] != 1 || num[1][2] || num[2][1]) return 0;
        return 1;
    }
    void clear(){
        memset(num, 0, sizeof(num));
    }
    matrix operator * (const matrix &b) const{
        matrix r;
        for(int i = 1; i <= 2; i++)
            for(int j = 1; j <= 2; j++)
                for(int k = 1; k <= 2; k++)
                    r.num[i][j] = (r.num[i][j] + num[i][k] * b.num[k][j]) % mod;
        return r;
    }
    matrix operator + (const matrix &b) const{
        matrix r;
        for(int i = 1; i <= 2; i++)
            for(int j = 1; j <= 2; j++)
                r.num[i][j] = (num[i][j] + b.num[i][j]) % mod;
        return r;
    }
    matrix operator ^ (ll p) const{
        matrix r, a;
        memcpy(a.num, num, sizeof(num));
        r.init();
        for(; p; p >>= 1, a = a * a)
            if(p & 1) r = r * a;
        return r;
    }
}f, A;

#define ls rt << 1
#define rs rt << 1 | 1

const ll N = 1e5 + 10;
ll n, m;
ll a[N];
matrix sum[N << 2], lazy[N << 2];

inline void pushup(ll rt){
    sum[rt] = sum[ls] + sum[rs];
}

inline void pushdown(ll l, ll r, ll rt){
    if(lazy[rt].empty()) return;
    ll mid = (l + r) >> 1;
    sum[ls] = sum[ls] * lazy[rt];
    sum[rs] = sum[rs] * lazy[rt];
    lazy[ls] = lazy[ls] * lazy[rt];
    lazy[rs] = lazy[rs] * lazy[rt];
    lazy[rt].clear();
    lazy[rt].init();
}

inline void build(ll l, ll r, ll rt){
    lazy[rt].init();
    if(l == r){
        sum[rt] = f * (A ^ (a[l] - 1));
        return;
    }
    ll mid = (l + r) >> 1;
    build(l, mid, ls);
    build(mid + 1, r, rs);
    pushup(rt);
}

inline void update(matrix x, ll L, ll R, ll l, ll r, ll rt){
    if(L <= l && r <= R){
        sum[rt] = sum[rt] * x;
        lazy[rt] = lazy[rt] * x;
        return;
    }
    pushdown(l, r, rt);
    ll mid = (l + r) >> 1;
    if(L <= mid) update(x, L, R, l, mid, ls);
    if(R > mid) update(x, L, R, mid + 1, r, rs);
    pushup(rt);
}

matrix query(ll L, ll R, ll l, ll r, ll rt){
    if(L <= l && r <= R)
        return sum[rt];
    pushdown(l, r, rt);
    ll mid = (l + r) >> 1;
    matrix res;
    if(L <= mid) res = res + query(L, R, l, mid, ls);
    if(R > mid) res = res + query(L, R, mid + 1, r, rs);
    return res;
}

signed main(){
    f.num[1][1] = f.num[1][2] = 1;
    A.num[1][1] = A.num[1][2] = A.num[2][1] = 1;
    ll n = read(), m = read();
    for(int i = 1; i <= n; i++)
        a[i] = read();
    build(1, n, 1);
    while(m--){
        ll op = read(), l = read(), r = read(), x;
        if(op == 1){
            x = read();
            update(A ^ x, l, r, 1, n, 1);
        }else
            printf("%lld\n", query(l, r, 1, n, 1).num[1][2] % mod);
    }
    return 0;
}

End