环鸽的数列

(7 mins to read)

区间加广义斐波那契数列大概有三种做法1.用特征方程求出通项,跑出根号的二次剩余后代入,变成区间加等比数列,线段树维护即可2.矩阵3.利用性质

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#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define fi first
#define se second
#define debug(x) cerr << #x << " " << x << '\n'
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pli = pair<ll,int>;
const int INF = 0x3f3f3f3f, N = 3e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 9, a = 276601605, b = 691504013, c = 308495997;
int sum[N];
int n, m, x[N];
inline void mul(int &a, int b) { a = 1ll*a*b%mod; }
inline int Mul(int a, int b) { return 1ll*a*b%mod; }
inline void add(int &a, int b) { a+=b; if(a>=mod) a-=mod; }
inline int Add(int a, int b) { a+=b; if(a>=mod) return a-mod; return a; }
inline void sub(int &a, int b) { a-=b; if(a<0) a+=mod; }
inline int Sub(int a, int b) { a-=b; if(a<0) return a+mod; return a; }
int powmod(int a, int b)
{
int ans = 1;
while(b)
{
if(b&1) ans = Mul(ans, a);
a = Mul(a, a);
b >>= 1;
}
return ans;
}
struct SegTree
{
#define ls (p<<1)
#define rs (p<<1|1)
int inv, pw[N];
struct seg
{
int l, r, sum, tag;
}t[N<<2];
void init(int q)
{
pw[0] = 1;
for(int i=1; i<=n; i++) pw[i] = Mul(pw[i-1], q)%mod;
inv = powmod(q-1, mod-2);
}
void setval(int p, int v)
{
add(t[p].sum, Mul(Mul(v, Sub(pw[t[p].r-t[p].l+1], 1)), inv));
add(t[p].tag, v);
}
void pushup(int p)
{
t[p].sum = Add(t[ls].sum, t[rs].sum);
}
void pushdown(int p)
{
setval(ls, t[p].tag); setval(rs, Mul(t[p].tag, pw[t[rs].l-t[ls].l]));
t[p].tag = 0;
}
void build(int p, int l, int r)
{
t[p].l = l, t[p].r = r;
if(l==r) return;
int mid = (l+r)>>1;
build(ls, l, mid); build(rs, mid+1, r);
pushup(p);
}
void upd(int p, int x, int y)
{
int l = t[p].l, r = t[p].r;
if(l>=x && r<=y)
{
setval(p, pw[l-x+1]);
return;
}
if(t[p].tag) pushdown(p);
int mid = (l+r)>>1;
if(x<=mid) upd(ls, x, y);
if(y>mid) upd(rs, x, y);
pushup(p);
}
int ask(int p, int x, int y)
{
int l = t[p].l, r = t[p].r;
if(l>=x && r<=y) return t[p].sum;
if(t[p].tag) pushdown(p);
int mid = (l+r)>>1, ans = 0;
if(x<=mid) add(ans, ask(ls, x, y));
if(y>mid) add(ans, ask(rs, x, y));
return ans;
}
#undef ls
#undef rs
}T1, T2;
int main()
{
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++) scanf("%d", x+i);
for(int i=1; i<=n; i++) sum[i] = Add(sum[i-1], x[i]);
T1.init(b); T2.init(c);
T1.build(1, 1, n); T2.build(1, 1, n);
while(m--)
{
int op, l, r;
scanf("%d%d%d", &op, &l, &r);
if(op==1)
{
T1.upd(1, l, r);
T2.upd(1, l, r);
}
else
{
int ans = Mul(Sub(T1.ask(1, l, r), T2.ask(1, l, r)), a);
add(ans, Sub(sum[r], sum[l-1]));
printf("%d\n", ans);
}
}
return 0;
}