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

最新推荐文章于 2020-12-18 15:18:01 发布

转载最新推荐文章于 2020-12-18 15:18:01 发布 · 242 阅读

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原文链接：http://www.cnblogs.com/LSTete/p/9533089.html

本文详细解析了一种利用线段树进行区间加斐波那契数列和区间求和的高效算法，通过巧妙转换斐波那契数列求和公式，实现对大规模数据的快速处理。

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;
}

转载于:https://www.cnblogs.com/LSTete/p/9533089.html