玄关求条(或者hack)，经典斐波那契线段树题，

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=3e5+10,mod=1e9+7;
ll fbi[N],sfbi[N];
int n,m;
ll a[N];
struct node{
	int l,r;
	ll tag1,tag2;
	ll sum;
}rt[4*N];
void build(int l,int r,int k){
	rt[k].l=l,rt[k].r=r,rt[k].tag1=rt[k].tag2=0;
	if(l==r){rt[k].sum=a[l];rt[k].sum%=mod;return;}
	int mid=(l+r)>>1;
	build(l,mid,k<<1),build(mid+1,r,k<<1|1);
	rt[k].sum=rt[k<<1].sum+rt[k<<1|1].sum;
	rt[k].sum%=mod;
}
void upsum(int k){
	rt[k].sum=(rt[k<<1].sum+rt[k<<1|1].sum)%mod;
}
void down(int k){

	rt[k<<1].tag1+=rt[k].tag1;
	rt[k<<1].tag2+=rt[k].tag2;
	int len=rt[k<<1].r-rt[k<<1].l+1;
	rt[k<<1].sum+=(rt[k].tag1*fbi[len]+rt[k].tag2*fbi[len+1]+mod-rt[k].tag2);
	rt[k<<1].sum%=mod;
	
	rt[k<<1].tag1%=mod;
	rt[k<<1].tag2%=mod;
	//upsum(k<<1);
	//下放左儿子标签 
	int pos=rt[k<<1|1].l-rt[k].l+1;
	ll adt1=(rt[k].tag1*fbi[pos-2]+rt[k].tag2*fbi[pos-1])%mod,
	adt2=(rt[k].tag1*fbi[pos-1]+rt[k].tag2*fbi[pos])%mod;
	//计算右儿子前两项的增量 
	rt[k<<1|1].tag1+=adt1;
	rt[k<<1|1].tag2+=adt2;
	len=rt[k<<1|1].r-rt[k<<1|1].l+1;
	rt[k<<1|1].sum+=(adt1*fbi[len]+adt2*fbi[len+1]+mod-adt2);
	rt[k<<1|1].sum%=mod;
	rt[k<<1|1].tag1%=mod;
	rt[k<<1|1].tag2%=mod;
	//下方右儿子标签 
	//rt[k<<1|1 sum+
	//upsum(k);
	rt[k].tag1=rt[k].tag2=0;
	//clear
	//rt[k].sum=rt[]
	
}
void update(int l,int r,int k){
	if(l<=rt[k].l&&rt[k].r<=r){
		rt[k].sum+=(sfbi[rt[k].r-l+1]-sfbi[rt[k].l-l]+mod);
		rt[k].sum%=mod;
		rt[k].tag1+=fbi[rt[k].l-l+1];
		rt[k].tag1%=mod;
		rt[k].tag2+=fbi[rt[k].l+1-l+1];
		rt[k].tag2%=mod;
		//upsum(k);
		return;
	}
	down(k);
	if(r>=rt[k<<1|1].l)update(l,r,k<<1|1);
	if(l<=rt[k<<1].r)update(l,r,k<<1);
	upsum(k);
}
ll que(int l,int r,int k){
	ll ans=0;
	if(l<=rt[k].l&&rt[k].r<=r){
		ans+=rt[k].sum;
		ans%=mod;
		return ans;
	}
	down(k);
	if(r>=rt[k<<1|1].l)ans+=que(l,r,k<<1|1),ans%=mod;
	if(l<=rt[k<<1].r)ans+=que(l,r,k<<1),ans%=mod;
	return ans;
}

int main(){
	cin>>n>>m;
	fbi[1]=fbi[2]=1;
	sfbi[1]=1,sfbi[2]=2;
	for(int i=3;i<=n+5;i++){
		fbi[i]=fbi[i-1]+fbi[i-2];
		sfbi[i]=(fbi[i]+sfbi[i-1])%mod;
		fbi[i]%=mod;
	}
	for(int i=1;i<=n;i++)
		cin>>a[i];
	build(1,n,1);
	for(int i=1;i<=m;i++){
		int opt,l,r;
		cin>>opt>>l>>r;
		if(opt==1){
			update(l,r,1);
		}
		if(opt==2){
			cout<<que(l,r,1)<<"\n";
		}
	}
	return 0; 
}