线段树之单点更新，区域求和

今天在coreforces上做的一题

E. DZY Loves Fibonacci Numbers

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. 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.
2. 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.

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

Note

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.

大意：用fib去加到每个节点，求区域的和

代码如下：

 
#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;

const int MAXNODE = 2097152; // == (1 << 21) - 1  depth:21
const int MAXS = 300003;
struct SOILD{
	int left,right; // 数组模拟满二叉树, 区间 [left,right]
	long value;
}Soldier[MAXNODE];
long fib[300005];
int father[MAXS]; // 存每一个士兵的父结点

void BuildTree(int i,int left,int right,int devalue){ // 初始值为0
	Soldier[i].left = left;
	Soldier[i].right = right;
	Soldier[i].value = devalue;
	if (left == right){
		father[left] = i; // 可以瞬间知道第i个士兵的父节点是哪个
		return;
	}
	BuildTree(i*2, left, (int)floor((right+left)/2.0), 0);
	BuildTree(i*2+1, (int)floor((right+left)/2.0)+1, right, 0);
}

void UpdataTree(int ri,int iKilln){ // 单点更新: 点p[i,i] 更新为iKilln (初始为0)
	if (ri == 1){
		Soldier[ri].value += iKilln;
		return;
	}
	Soldier[ri].value += iKilln; // iKilln 设为全局变量应该快一点
	UpdataTree(ri/2,iKilln);
}

long res;
void Query(int i,int a,int b){ // [a,b]区间的和
	if (a == Soldier[i].left && b == Soldier[i].right){
		res += Soldier[i].value;
		return ;
	}
	if (a <= Soldier[i*2].right){ // 左端点小于右端点（几何含义：区间横跨过中点） 则区间需要分割
		if (b <= Soldier[i*2].right){
			Query(i*2, a, b); // 左边全包含
		}else{
			Query(i*2, a, Soldier[i*2].right); // 横跨取左
		}
	}
	if (b >= Soldier[i*2+1].left){
		if (a >= Soldier[i*2+1].left){
			Query(i*2+1, a, b); // 右边全包含
		}else{
			Query(i*2+1, Soldier[i*2+1].left, b); // 横跨取右
		}
	}
}

int main(){
	int n_s,n_q;
	long iKilln,a,b;
	cin >> n_s >> n_q;
	BuildTree(1,1,n_s,0);
	fib[1]=1;fib[2]=1;
	for(int k2=3;k2<300005;k2++)
	{
		fib[k2]=fib[k2-1]+fib[k2-2];
	}
	for (int ii = 1 ;ii <= n_s; ii++){ // 建树
		scanf("%d",&iKilln);	// iKilln 为第 i 个增加的杀敌数
		UpdataTree(father[ii],iKilln); // 瞬间找到士兵的结点序号
	}
	int order;
	for (int i = 1; i <= n_q; i++){
		scanf("%d%d%d",&order,&a,&b); // [a,b] ; a + kill b 	
		if (order==2){
			res = 0;
			Query(1,a,b);
			printf("%d\n", res);
		}else
		{
			for(int kk=a;kk<=b;kk++)
		UpdataTree(father[kk],fib[kk-a+1]);
		}
		//for(int ttt=1;ttt<=n_s;ttt++)
	//	printf("%d;;",Soldier[father[ttt]].value);
	//	printf("\n");
	
	}
	return 0;
}

 要取 long long/__int64；

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