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Description

? 看题戳我 给你一个序列，要求支持区间加斐波那契数列和区间求和。\(~n \leq 3 \times 10 ^ 5, ~fib_1 = fib_2 = 1~\).

Solution

? 先来考虑一段斐波那契数列如何快速求和，根据性质有
\[ \begin {align} fib_n &= fib_{n - 1} + fib_{n - 2} \ &= fib_ {n - 2} + fib_{n - 3} + fib_{n - 2} \ &= fib_{n - 3} + fib_{n - 4} + fib_{n - 3} + fib_{n - 2} \ &= \dots \ &= fib_2 + \sum_{i = 1}^{n - 2} {fib_i} \end {align} \]
? 可以发现这里有个\(~\sum_{i = 1} ^ {n - 2} {fib_i}\)，转换一下就是\(~\sum_{i = 1} ^ {n}fib_i = fib_{n + 2} - fib_2\).而两个斐波那契数列对应项加起来之后还是一个类斐波那契数列，记为\(~S_i\)，设这个类斐波那契数列的起始项\(S_1 = a, S_2 = b\)，显然有\(~S_i = a \times fib_{i - 2} + b \times fib_{i - 1}\).那么对于一段类斐波那契数列的求和，我们只要记起始的两项和这段数列的长度即可。现在可以用简单的线段树区间加来维护了，\(~PushDown~\)操作有一点细节，注意要分开算区间的前两项。具体看代码。。

Code

#include<bits/stdc++.h>
#define For(i, j, k) for(int i = j; i <= k; ++i)
#define Forr(i, j, k) for(int i = j; i >= k; --i)
using namespace std;

inline int read() {
    int x = 0, p = 1; char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == ‘-‘) p = -1;
    for(; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x *= p;
}

inline void File() {
    freopen("cf446c.in", "r", stdin);
    freopen("cf446c.out", "w", stdout);
}

const int N = 3e5 + 10, mod = 1e9 + 9;
int n, m, fib[N];

inline int add(int a, int b) { return (a += b) >= mod ? a - mod : a; }

namespace SGT {
#define lc (rt << 1)
#define rc (rt << 1 | 1)
#define mid (l + r >> 1)
#define lson lc, l, mid
#define rson rc, mid + 1, r

    int tr[N << 2], t1[N << 2], t2[N << 2];
    
    inline void pushup(int rt) { tr[rt] = (tr[lc] + tr[rc]) % mod; }

    inline int S(int a, int b, int x) {
        return x == 1 ? a : (x == 2 ? b : (1ll * a * fib[x - 2] + 1ll * b * fib[x - 1]) % mod); 
    } 

    inline int sum(int a, int b, int x) {
        return x == 1 ? a : (x == 2 ? add(a, b) : (S(a, b, x + 2) - b + mod) % mod);    
    }

    inline void pushdown(int rt, int l, int r) {
        if (t1[rt]) {
            t1[lc] = add(t1[lc], t1[rt]), t2[lc] = add(t2[lc], t2[rt]);
            tr[lc] = add(tr[lc], sum(t1[rt], t2[rt], mid - l + 1));     
            int T1 = S(t1[rt], t2[rt], mid - l + 2), T2 = S(t1[rt], t2[rt], mid - l + 3);
            t1[rc] = add(t1[rc], T1), t2[rc] = add(t2[rc], T2);
            tr[rc] = add(tr[rc], sum(T1, T2, r - mid));
            t1[rt] = t2[rt] = 0;
        }
    }

    inline void build(int rt, int l, int r) {
        if (l == r) tr[rt] = read();
        else build(lson), build(rson), pushup(rt);
    }

    inline void update(int rt, int l, int r, int L, int R) {
        if (L <= l && r <= R) {
            tr[rt] = add(tr[rt], sum(fib[l - L + 1], fib[l - L + 2], r - l + 1));
            t1[rt] = add(t1[rt], fib[l - L + 1]); t2[rt] = add(t2[rt], fib[l - L + 2]); 
            return ;
        }   
        pushdown(rt, l, r);
        if (L <= mid) update(lson, L, R);
        if (R > mid) update(rson, L, R);
        pushup(rt);
    }

    inline int query(int rt, int l, int r, int L, int R) {
        if (L <= l && r <= R) return tr[rt];
        pushdown(rt, l, r); int res = 0;
        if (L <= mid) res = add(res, query(lson, L, R));
        if (R > mid) res = add(res, query(rson, L, R));
        return pushup(rt), res;
    }

#undef lc
#undef rc
#undef mid
#undef lson
#undef rson
}

int main() {
    File();
    n = read(), m = read();
    fib[1] = fib[2] = 1; 
    For(i, 3, n + 5) fib[i] = (fib[i - 1] + fib[i - 2]) % mod;

    using namespace SGT;
    build(1, 1, n);
    while (m --) {
        int opt = read(), l = read(), r = read();
        opt == 1 ? update(1, 1, n, l, r), 1 : printf("%d\n", query(1, 1, n, l, r)), 1;
    }

    return 0;
}

【CF446C】DZY Loves Fibonacci Numbers （线段树 + 斐波那契数列）

标签：支持   ble   就是   har   .com   oid   快速   stdin   cpp   

原文地址：https://www.cnblogs.com/LSTete/p/9533089.html