noip2017集训测试赛（四）Problem A: fibonacci

题目大意

给你一个序列\(a_1, a_2, ..., a_n\). 我们令函数\(f(n)\)表示斐波那契数列第\(n\)项的值. 总共\(m\)个操作, 分为以下两种:

• 将\(x \in [L, R]\)中的所有\(a_x\)加上一个数\(k\);
• 询问\(\sum_{x \in [L, R]}f(a_x)\)

\(n \le 10^5\)

\(m \le 10^5\)

Solution

我们靠考虑斐波那契数列的转移矩阵:

\[[f_{a + b - 1} \ f_{a + b}] = [f_{a - 1} \ f_a] \left[ \begin{array}{} 0 \ 1 \\ 1 \ 1 \end{array} \right]^b
\]

同时我们发现若干个斐波那契数之和也满足这个关系.

因此我们用线段树维护每个区间的\(\sum_{L \le i \le R} f(a_i)\)以及\(\sum_{L \le i \le R} f(a_{i - 1})\), 区间修改查询即可.

注意作为常数优化, 我们要预处理出转移矩阵的\(2^n\)次幂. 否则会TLE.

话说拿这种小技巧, 把正确复杂度的代码卡到30分, 是不是也太丧心病狂了!!!

#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>

namespace Zeonfai
{
    inline long long getInt()
    {
        long long a = 0, sgn = 1; char c;
        while(! isdigit(c = getchar())) if(c == '0') sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const long long N = (long long)1e5, MOD = (long long)1e9 + 7;
long long n;
struct matrix
{
    long long a[2][2];
    inline matrix operator *(const matrix &A)
    {
        matrix res; memset(res.a, 0, sizeof(res.a));
        for(long long i = 0; i < 2; ++ i) for(long long j = 0; j < 2; ++ j) for(long long k = 0; k < 2; ++ k)
            res.a[i][j] = (res.a[i][j] + (long long)a[i][k] * A.a[k][j] % MOD) % MOD;
        return res;
    }
}trans, ptt[64];
inline matrix power(long long x)
{
    matrix res; memset(res.a, 0, sizeof(res.a)); res.a[0][0] = res.a[1][1] = 1;
    if(x < 0) return res;
    long long p = 0;
    for(; x; x >>= 1, ++ p)
        if(x & 1) res = res * ptt[p];
    return res;
}
struct segmentTree
{
    struct node
    {
        long long cur, lst;
        long long tg; // 注意到tg的值叠加后可能超过Long long
        inline node() {cur = 0; lst = 1; tg = 0;}
    }nd[N << 2];
    inline void pushDown(long long _u)
    {
        long long x = nd[_u].tg; long long u = _u << 1;
        matrix res = power(x);
        long long a = ((long long)nd[u].lst * res.a[0][0] % MOD + (long long)nd[u].cur * res.a[1][0] % MOD) % MOD,
            b = ((long long)nd[u].lst * res.a[0][1] % MOD + (long long)nd[u].cur * res.a[1][1] % MOD) % MOD;
        nd[u].lst = a; nd[u].cur = b;
        nd[u].tg += x;
        u = _u << 1 | 1;
        a = ((long long)nd[u].lst * res.a[0][0] % MOD + (long long)nd[u].cur * res.a[1][0] % MOD) % MOD;
        b = ((long long)nd[u].lst * res.a[0][1] % MOD + (long long)nd[u].cur * res.a[1][1] % MOD) % MOD;
        nd[u].lst = a; nd[u].cur = b;
        nd[u].tg += x;
        nd[_u].tg = 0;
    }
    void modify(long long u, long long curL, long long curR, long long L, long long R, long long x)
    {
        if(curL >= L && curR <= R)
        {
            nd[u].tg += x;
            matrix res = power(x);
            long long a = ((long long)nd[u].lst * res.a[0][0] % MOD + (long long)nd[u].cur * res.a[1][0] % MOD) % MOD,
                b = ((long long)nd[u].lst * res.a[0][1] % MOD + (long long)nd[u].cur * res.a[1][1] % MOD) % MOD;
            nd[u].lst = a; nd[u].cur = b;
            return;
        }
        pushDown(u);
        long long mid = curL + curR >> 1;
        if(L <= mid) modify(u << 1, curL, mid, L, R, x);
        if(R > mid) modify(u << 1 | 1, mid + 1, curR, L, R, x);
        nd[u].cur = (nd[u << 1].cur + nd[u << 1 | 1].cur) % MOD;
        nd[u].lst = (nd[u << 1].lst + nd[u << 1 | 1].lst) % MOD;
    }
    inline void modify(long long L, long long R, long long x) {modify(1, 1, n, L, R, x);}
    long long query(long long u, long long curL, long long curR, long long L, long long R)
    {
        if(curL >= L && curR <= R) return nd[u].cur;
        pushDown(u);
        long long mid = curL + curR >> 1, res = 0;
        if(L <= mid) res = query(u << 1, curL, mid, L, R);
        if(R > mid) res = (res + query(u << 1 | 1, mid + 1, curR, L, R)) % MOD;
        return res;
    }
    inline long long query(long long L, long long R) {return query(1, 1, n, L, R);}
}seg;
int main()
{

#ifndef ONLINE_JUDGE

    freopen("fibonacci.in", "r", stdin);
    freopen("_fibonacci.out", "w", stdout);

#endif

    using namespace Zeonfai;
    trans.a[0][0] = 0; trans.a[0][1] = trans.a[1][0] = trans.a[1][1] = 1;
    ptt[0] = trans;
    for(long long i = 1; i < 64; ++ i) ptt[i] = ptt[i - 1] * ptt[i - 1];
    n = getInt(); long long m = getInt();
    for(long long i = 1; i <= n; ++ i) seg.modify(i, i, getInt());
    for(long long i = 0; i < m; ++ i)
    {
        long long opt = getInt(), L = getInt(), R = getInt();
        if(opt == 1)
        {
            long long x = getInt();
            seg.modify(L, R, x);
        }
        else if(opt == 2) printf("%d\n", seg.query(L, R));
    }
}