[CF633H] Fibonacci-ish II
多个区间和去重交给莫队。
用线段树维护 \(\begin{bmatrix} fib_{i-1} \times val, fib_i \times val \end{bmatrix}\)。
具体和线段树的区间和差不多,上传时左右儿子矩阵相加,下传时左右儿子矩阵都乘上懒标记。
斐波那契数列下一项的矩阵是 \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\),插入新数时区间乘这个矩阵就行。
区间除要用刚才矩阵的逆矩阵 \(\begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix}\),现在就变成乘法了。
卡常小技巧:
用 unsigned short,矩阵乘法展开,离散化。
#include <bits/stdc++.h>
#define int unsigned short
#define ls p << 1
#define rs p << 1 | 1
#define Ad tr[p].ad
#define mid (l + r >> 1)
using namespace std;
inline signed read()
{
signed f = 0, ans = 0;
char c = getchar();
while (!isdigit(c))
f |= c == '-', c = getchar();
while (isdigit(c))
ans = (ans << 3) + (ans << 1) + c - 48, c = getchar();
return f ? -ans : ans;
}
void write(int x)
{
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
const signed N = 3e4 + 5, M = N << 2;
signed a[N], b[N];
int n, m, sq, si, mod;
int fib[N], res[N], mp[N], id[N];
struct que
{
int l, r, id;
bool operator<(const que &x) const
{
if (l / sq != x.l / sq)
return l < x.l;
return l / sq & 1 ? r < x.r : r > x.r;
}
} q[N];
template <typename T>
inline T Mod(T &&x) { return x >= mod ? x -= mod : x; }
struct Matrix
{
int h, w, a[2][2];
Matrix() : Matrix(1, 2) {}
Matrix(int H, int W) : h(H), w(W)
{
for (int i = 0; i < h; ++i)
for (int j = 0; j < w; ++j)
a[i][j] = 0;
}
Matrix(short id)
{
if (id == 0)
assert(0);
else
{
h = w = 2;
if (id > 0)
a[0][1] = a[1][0] = a[1][1] = 1, a[0][0] = 0;
else
a[1][0] = a[0][1] = 1, a[0][0] = mod - 1, a[1][1] = 0;
}
}
Matrix operator+(const Matrix &rhs) const
{
Matrix ans(*this);
Mod(ans.a[0][0] += rhs.a[0][0]);
Mod(ans.a[0][1] += rhs.a[0][1]);
return ans;
}
Matrix operator*(const Matrix &rhs) const
{
Matrix ans(h, rhs.w);
if (h == 1 && w == 2)
{
ans.a[0][0] = Mod(1 * a[0][0] * rhs.a[0][0] % mod + 1 * a[0][1] * rhs.a[1][0] % mod);
ans.a[0][1] = Mod(1 * a[0][0] * rhs.a[0][1] % mod + 1 * a[0][1] * rhs.a[1][1] % mod);
}
else
{
ans.a[0][0] = Mod(1 * a[0][0] * rhs.a[0][0] % mod + 1 * a[0][1] * rhs.a[1][0] % mod);
ans.a[0][1] = Mod(1 * a[0][0] * rhs.a[0][1] % mod + 1 * a[0][1] * rhs.a[1][1] % mod);
ans.a[1][0] = Mod(1 * a[1][0] * rhs.a[0][0] % mod + 1 * a[1][1] * rhs.a[1][0] % mod);
ans.a[1][1] = Mod(1 * a[1][0] * rhs.a[0][1] % mod + 1 * a[1][1] * rhs.a[1][1] % mod);
}
return ans;
}
} u[N], ru[N];
struct Tree
{
struct tree
{
Matrix v;
int sum;
short ad;
} tr[M];
void down(signed p)
{
if (Ad > 0)
{
tr[ls].v = tr[ls].v * u[Ad], tr[ls].ad += Ad;
tr[rs].v = tr[rs].v * u[Ad], tr[rs].ad += Ad;
Ad = 0;
}
else if (Ad < 0)
{
tr[ls].v = tr[ls].v * ru[-Ad], tr[ls].ad += Ad;
tr[rs].v = tr[rs].v * ru[-Ad], tr[rs].ad += Ad;
Ad = 0;
}
}
void update(int p) { tr[p].v = tr[ls].v + tr[rs].v; }
void insert(int l, int r, int x, signed p, int rk)
{
++tr[p].sum;
if (l == r)
{
tr[p].v = Matrix(1, 2);
tr[p].v.a[0][0] = 1 * fib[rk - 1] * b[x] % mod;
tr[p].v.a[0][1] = 1 * fib[rk] * b[x] % mod;
return;
}
down(p);
if (mid >= x)
insert(l, mid, x, ls, rk);
else
insert(mid + 1, r, x, rs, rk);
update(p);
}
void insert(int x, int rk) { insert(1, si, x, 1, rk); }
void remove(int l, int r, int x, signed p)
{
--tr[p].sum;
if (l == r)
{
tr[p].v = Matrix();
return;
}
down(p);
if (mid >= x)
remove(l, mid, x, ls);
else
remove(mid + 1, r, x, rs);
update(p);
}
void remove(int x) { remove(1, si, x, 1); }
void multiply(int l, int r, int s, int t, signed p, Matrix &v, short w)
{
if (l >= s && r <= t)
{
tr[p].v = tr[p].v * v, tr[p].ad += w;
return;
}
down(p);
if (mid >= s)
multiply(l, mid, s, t, ls, v, w);
if (mid < t)
multiply(mid + 1, r, s, t, rs, v, w);
update(p);
}
void multiply(int s, int t, Matrix &v, short w) { multiply(1, si, s, t, 1, v, w); }
int ask_ans() { return tr[1].v.a[0][1]; }
int rank(int l, int r, int s, int t, signed p)
{
if (l >= s && r <= t)
return tr[p].sum;
down(p);
int ans = 0;
if (mid >= s)
ans = rank(l, mid, s, t, ls);
if (mid < t)
ans += rank(mid + 1, r, s, t, rs);
return ans;
}
int rank(int x) { return rank(1, si, 1, x, 1); }
} tr;
inline void insert(int i)
{
if (++mp[id[i]] == 1)
{
tr.insert(id[i], tr.rank(id[i]) + 1);
tr.multiply(id[i] + 1, n, u[1], 1);
}
}
inline void remove(int i)
{
if (--mp[id[i]] == 0)
{
tr.remove(id[i]);
tr.multiply(id[i] + 1, n, ru[1], -1);
}
}
signed main()
{
// freopen("chk.in", "r", stdin);
// freopen("chk.out", "w", stdout);
n = read(), sq = sqrt(n);
mod = read();
for (int i = 1; i <= n; ++i)
a[i] = read();
copy(a + 1, a + n + 1, b + 1);
sort(b + 1, b + n + 1);
si = unique(b + 1, b + n + 1) - b;
for (int i = 1; i <= n; ++i)
id[i] = lower_bound(b + 1, b + si, a[i]) - b;
for (int i = 1; i <= n; ++i)
b[i] = b[i] % mod;
fib[1] = fib[2] = 1;
for (int i = 3; i <= n; ++i)
fib[i] = Mod(fib[i - 1] + fib[i - 2]);
u[1] = Matrix(1), ru[1] = Matrix(-1);
for (int i = 2; i <= n; ++i)
u[i] = u[i - 1] * u[1], ru[i] = ru[i - 1] * ru[1];
m = read();
for (int i = 1; i <= m; ++i)
q[i].l = read(), q[i].r = read(), q[i].id = i;
sort(q + 1, q + m + 1);
int l = 1, r = 0;
for (int i = 1; i <= m; ++i)
{
while (l > q[i].l)
insert(--l);
while (r < q[i].r)
insert(++r);
while (l < q[i].l)
remove(l++);
while (r > q[i].r)
remove(r--);
res[q[i].id] = tr.ask_ans();
}
for (int i = 1; i <= m; ++i)
write(res[i]), putchar('\n');
return 0;
}

浙公网安备 33010602011771号