Codeforces Round #FF (Div. 2)E（线段树成段更新）

最新推荐文章于 2022-12-16 16:19:21 发布

cq_phqg

最新推荐文章于 2022-12-16 16:19:21 发布

阅读量731

点赞数

CC 4.0 BY-SA版权

分类专栏：数据结构_线段树文章标签： ACM codeforces

本文链接：https://blog.youkuaiyun.com/cq_phqg/article/details/38980265

数据结构_线段树专栏收录该内容

12 篇文章

订阅专栏

博客围绕斐波那契数列在数组查询中的应用展开。给定含n个整数的数组及m个查询，查询分两种类型，一是对指定区间元素加上对应斐波那契数，二是输出指定区间元素和取模结果。需根据输入数据处理并回复所有查询。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2).

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 $\text{[math]}$ modulo 1000000009 (109 + 9).

Help DZY reply to all the queries.

Input

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.

Output

For each query of the second type, print the value of the sum on a single line.

Sample test(s)

input
4 4
1 2 3 4
1 1 4
2 1 4
1 2 4
2 1 3

output
17
12

题意：支持两种操作，1 l r 将 ai 加上第i-l+1项Fibonacci数，l<=i<=r ，2 l r 求区间和

思路：根据Fibonacci数列的性质，只要知道前两项，那么可以O（1）得到后面任意项的Fibonacci数以及任意前多少项的和，可以用线段树维护每个线段的和以及每段的前两

个Fibonacci数，延迟更新的是每个线段的前两个Fibonacci数（直接累加）

#include 
#include 
#include 
#include 
using namespace std;
#define lson l,m,o<<1
#define rson m+1,r,o<<1|1
#define maxn 300010
#define mod 1000000009

typedef long long ll;
ll a[maxn<<2];
ll b[maxn<<2];
ll sum[maxn<<2];
ll fibo[maxn];
ll sfibo[maxn];
int n,m;

ll addit(ll a,ll x)
{
    a+=x;
    if(a>=mod)a-=mod;
    return a;
}

void pushup(int o)
{
    sum[o]=addit(sum[o<<1],sum[o<<1|1]);
}

void build(int l,int r,int o)
{
    a[o]=b[o]=0;
    if(l==r){
        cin>>sum[o];
        return;
    }
    int m=l+r>>1;
    build(lson);
    build(rson);
    pushup(o);
}

ll chul(ll a,ll b,int n)
{
    if(n==1)return a;
    if(n==2)return b;
    return addit( fibo[n-2] * a % mod , fibo[n-1] * b % mod );
}

ll chuli(ll a,ll b,int n)
{
    if(n==1)return a;
    if(n==2)return addit(a,b);
    return addit( addit(sfibo[n-2],1) * a % mod , sfibo[n-1] * b % mod );
}

void pushdown(int o,int n)
{
    if(a[o])
    {
        int LL=n/2+n%2;
        int rr=n-LL;

        a[o<<1]=addit(a[o<<1],a[o]);
        b[o<<1]=addit(b[o<<1],b[o]);

        sum[o<<1]=addit( sum[o<<1] , chuli(a[o],b[o],LL) );

        ll x=chul(a[o],b[o],LL+1);
        ll y=chul(a[o],b[o],LL+2);

        a[o<<1|1]=addit(a[o<<1|1],x);
        b[o<<1|1]=addit(b[o<<1|1],y);

        sum[o<<1|1]=addit( sum[o<<1|1] , chuli(x,y,rr) );

        a[o]=b[o]=0;
    }
}

void update(int l,int r,int o,int L,int R)
{
    if(L<=l&&r<=R){
        ll aa=chul(1,1,l-L+1);
        ll bb=chul(1,1,l-L+2);
        a[o]=addit(a[o],aa);
        b[o]=addit(b[o],bb);
        sum[o]=addit(sum[o], chuli(aa,bb,r-l+1) );
        return;
    }
    pushdown(o,r-l+1);
    int m=l+r>>1;
    if(L<=m)update(lson,L,R);
    if(m>1;
    ll ans=0;
    pushdown(o,r-l+1);
    if(L<=m)ans=addit(ans,query(lson,L,R));
    if(m