參考：http://www.cnblogs.com/chanme/p/3843859.html
然后我看到在别人的AC的方法里还有这么一种神方法，他预先设定了一个阈值K,当当前的更新操作数j<K的时候，它就用一个类似于树状数组段更的方法，用一个 d数组去存内容，譬如它要在区间 [3,6]上加一段fibonacci
原来：
id 0 1 2 3 4 5 6 7 8 9 10
d 0 0 0 0 0 0 0 0 0 0 0
更新:
id 0 1 2 3 4 5 6 7 8 9 10
d 0 0 0 1 0 0 0 -5 -3 0 0
我们能够发现，当利用 d[i]=d[i]+d[i-1]+d[i-2] i由小到大更新后就会得到
id 0 1 2 3 4 5 6 7 8 9 10
d 0 0 0 1 1 2 3 0 0 0 0
所以对于[L,R]上加一段操作，我们能够d[L]+=1; d[R+1]-=f[R-L+2],d[R+2]=f[R-L+1];
所以假如我更新了m次，我每次更新的复杂度是O(1),我把全部数求出来一次的复杂度是O(n)
然后他的算法是这种，j<K的时候更新的时候依照上述方法更新，询问的话则是将询问区间和更新的区间求交，把交的和加上去。
当j==K的时候，利用d还原出新的a，把当前的区间清0.
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: a1, a2, ..., an. Moreover, there are mqueries, each query has one of the two types:
- Format of the query "1 l r". In reply to the query, you need to add Fi - l + 1 to each element ai, where l ≤ i ≤ r.
-
Format of the query "2 l r". In reply to the query you should output the value of
modulo 1000000009 (109 + 9).
Help DZY reply to all the queries.
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 300000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — initial array a.
Then, m lines follow. A single line describes a single query in the format given in the statement. It is guaranteed that for each query inequality 1 ≤ l ≤ r ≤ n holds.
For each query of the second type, print the value of the sum on a single line.
4 4 1 2 3 4 1 1 4 2 1 4 1 2 4 2 1 3
17 12
After the first query, a = [2, 3, 5, 7].
For the second query, sum = 2 + 3 + 5 + 7 = 17.
After the third query, a = [2, 4, 6, 9].
For the fourth query, sum = 2 + 4 + 6 = 12.
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=330000;
const long long int mod=1000000009LL;
typedef long long int LL;
int n,m,k;
LL a[maxn],sum[maxn],fib[maxn],gib[maxn];
LL d[maxn]; int L[1100],R[1100];
void init()
{
fib[1]=1LL,fib[2]=1LL;
gib[1]=1LL,gib[2]=2LL;
for(int i=3;i<=n+50;i++)
{
fib[i]=(fib[i-1]+fib[i-2])%mod;
gib[i]=(gib[i-1]+fib[i])%mod;
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%I64d",a+i);
sum[i]=sum[i-1]+a[i];
}
init();
while(m--)
{
int c,l,r;
scanf("%d%d%d",&c,&l,&r);
if(c==1)
{
L[k]=l,R[k]=r; k++;
d[l]=(d[l]+1LL)%mod;
d[r+1]=(d[r+1]-fib[r-l+2]+mod)%mod;
d[r+2]=(d[r+2]-fib[r-l+1]+mod)%mod;
if(k>=1000)
{
for(int i=1;i<=n;i++)
{
if(i>=2)
d[i]=((d[i]+d[i-1])%mod+d[i-2])%mod;
a[i]=(a[i]+d[i])%mod;
sum[i]=sum[i-1]+a[i];
}
memset(d,0,sizeof(d)); k=0;
}
}
else if(c==2)
{
LL ret=0;
for(int i=0;i<k;i++)
{
int _left=max(l,L[i]);
int _right=min(r,R[i]);
if(_left>_right) continue;
int s=_left-L[i]+1,e=_right-L[i]+1;
ret=((ret+gib[e])%mod-gib[s-1]+mod)%mod;
}
ret=((ret+sum[r])%mod-sum[l-1]+mod)%mod;
printf("%I64d
",(ret+mod)%mod);
}
}
return 0;
}