CF446C DZY Loves Fibonacci Numbers 线段树 + 数学

Posted 2020-11-30 guangheli

code: 

#include <bits/stdc++.h> 
#define N 400004   
#define LL long long   
#define lson now<<1 
#define rson now<<1|1 
#define setIO(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)  
using namespace std;  
const LL mod=1000000009;            
int n,m;                     
LL fib[N<<1],sum[N<<1];    
struct node 
{
	LL f1,f2,sum;   
	int l,r,len;      
}t[N<<2];  
void build(int l,int r,int now) 
{ 
	t[now].l=l; 
	t[now].r=r; 
	t[now].len=r-l+1;  
	if(l==r) return ;  
	int mid=(l+r)>>1;      
	if(l<=mid)    build(l,mid,lson); 
	if(r>mid)     build(mid+1,r,rson);      
}  
void mark(int now,LL f1,LL f2) 
{    
	(t[now].f1+=f1)%=mod; 
	(t[now].f2+=f2)%=mod;        
	(t[now].sum+=f1*fib[t[now].len]%mod+f2*fib[t[now].len+1]%mod-f2+mod)%=mod;        
}   
void pushup(int now) 
{
	t[now].sum=(t[lson].sum+t[rson].sum)%mod;               
} 
void pushdown(int now) 
{
	if(t[now].f1==0&&t[now].f2==0) return;   
	int mid=(t[now].l+t[now].r)>>1;             
	mark(lson,t[now].f1,t[now].f2);    
	if(t[now].r>mid)   
	mark(rson,t[now].f1*fib[t[lson].len-1]%mod+t[now].f2*fib[t[lson].len]%mod,t[now].f1*fib[t[lson].len]%mod+t[now].f2*fib[t[lson].len+1]%mod);     
	t[now].f1=t[now].f2=0;    
}
void update(int l,int r,int now,int L,int R) 
{ 
	if(l>=L&&r<=R) 
	{            
		mark(now,fib[l-L+1],fib[l-L+2]);           
		return;       
	}    
	pushdown(now); 
	int mid=(l+r)>>1;  
	if(L<=mid)     update(l,mid,lson,L,R); 
	if(R>mid)      update(mid+1,r,rson,L,R);    
	pushup(now);      
}    
LL query(int l,int r,int now,int L,int R) 
{ 
	if(l>=L&&r<=R)     
	{    
		return t[now].sum;   
	}
	pushdown(now); 
	int mid=(l+r)>>1;   
	LL re=0ll;  
	if(L<=mid)    re+=query(l,mid,lson,L,R);   
	if(R>mid)     re+=query(mid+1,r,rson,L,R);   
	return re%mod; 
}
int main() 
{ 
	// setIO("input"); 
	int i,j; 
	scanf("%d%d",&n,&m); 
	fib[1]=fib[2]=1; 
	for(i=3;i<N;++i)    fib[i]=(fib[i-1]+fib[i-2])%mod;     
	for(i=1;i<=n;++i)    scanf("%lld",&sum[i]), (sum[i]+=sum[i-1])%=mod;    
	build(1,n,1);      
	for(i=1;i<=m;++i) 
	{   
		int opt,l,r;     
		scanf("%d%d%d",&opt,&l,&r);        
		if(opt==1) update(1,n,1,l,r); 
		else printf("%lld
",(query(1,n,1,l,r)+sum[r]-sum[l-1]+mod*2)%mod);    
	}    
	return 0;   
}