Codeforces Round #FF (Div. 2) E. Dzy Loves Fibonacci Numbers (Fibonacci theorem + line tree)

/*
Make full use of the two theorems of the Fibonacci series:
① definition f[1] = A, f[2] = b, f[n] = f[n-1] + f[n-2] (n≥3).
F[n] = b * fib[n-1] + A * fib[n-2] (n≥3), where Fib[i] is the first term of the Fibonacci sequence.
② definition f[1] = A, f[2] = b, f[n] = f[n-1] + f[n-2] (n≥3).
F[1] + f[2] + ... + f[n] = f[n + 2]-B
There is also a fact, that is, two of the above definitions of the sequence, add, still conform to f[n] = f[n-1] + f[n-2] recursive formula.
Using these two theorems, the sequence is maintained with the segment tree, and each node of the segment tree records what the first two items of this paragraph are, and the Fibonacci sequence is preprocessed.
, O (1) is able to calculate the number of the middle of each node, and the amount of each node.

*/

#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream>using namespace std; #define MAXN 300005#define Lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longconst ll mod=100000000 9;ll fibo[maxn],sum[maxn<<2],a[maxn<<2],b[maxn<<2];//Play table Fibonacci that tangent sequence void Get_fibo () {fibo[1]=fibo[2]=1    ; for (int i=3;i<maxn;i++) fibo[i]= (fibo[i-1]+fibo[i-2])%mod;}    According to the formula Two get the nth number of the meager series ll GET_FB (int n,ll a1,ll B1) {if (n==1) return a1%mod;    else if (n==2) return b1%mod; else return (b1*fibo[n-1]+a1*fibo[n-2])%mod;} void push_up (int rt) {sum[rt]= (sum[rt<<1]+sum[rt<<1|1])%mod;}    void Push_down (int rt,int l,int r) {int mid= (L+R) >>1;    A[rt<<1]= (A[rt<<1]+a[rt])%mod;    B[rt<<1]= (b[rt]+b[rt<<1])%mod;    A[rt<<1|1]= (A[RT&LT;&LT;1|1]+GET_FB (MID+1-L+1,A[RT],B[RT))%mod;    B[rt<<1|1]= (B[RT&LT;&LT;1|1]+GET_FB (MID+1-L+2,A[RT],B[RT))%mod; ll k1= (GET_FB (Mid-l+1+2,a[rt],b[rt])-b[Rt]+mod)%mod;      Record the first half and LL k2= (GET_FB (R-L+1+2,A[RT],B[RT))-b[rt]-k1+mod)%mod;    Record the second half of the and sum[rt<<1]= (SUM[RT&LT;&LT;1]+K1)%mod;    sum[rt<<1|1]= (SUM[RT&LT;&LT;1|1]+K2)%mod; a[rt]=b[rt]=0;}    void build (int l,int R,int RT) {a[rt]=b[rt]=0;    if (l==r) {scanf ("%d", &sum[rt]); return;    } int m= (L+R) >>1; Build (Lson);    Build (Rson); PUSH_UP (RT);}        void Update (LL l,ll R,LL l,ll r,ll RT) {if (l<=l&&r>=r) {a[rt]= (a[rt]+fibo[l-l+1])%mod;        B[rt]= (b[rt]+fibo[l-l+2])%mod;        Sum[rt]= (SUM[RT]+GET_FB (r-l+1+2,fibo[l-l+1],fibo[l-l+2])-fibo[l-l+2]+mod)%mod;    Return    } push_down (Rt,l,r);    ll m= (l+r) >>1;    if (l<=m) update (L,r,lson);    if (r>m) update (L,r,rson); PUSH_UP (RT);}    ll query (ll l,ll R,LL l,ll r,ll RT) {if (l<=l&&r>=r) return sum[rt]%mod;    Push_down (RT,L,R); ll m= (l+r) >>1;    ll ret=0;    if (l<=m) ret+=query (L,r,lson);    if (r>m) ret+=query (L,r,rson); RetUrn Ret%mod;}    int main () {ll i,j,n,m,a1,b1,c1;    Get_fibo ();    cin>>n>>m;    Build (1,n,1);        while (m--) {scanf ("%lld%lld%lld", &AMP;A1,&AMP;B1,&AMP;C1);        if (a1==1) update (b1,c1,1,n,1);    else printf ("%lld\n", (Query (b1,c1,1,n,1) +mod)%mod); } return 0;}

Codeforces Round #FF (Div. 2) E. Dzy Loves Fibonacci Numbers (Fibonacci theorem + line tree)