牛客多校第十场 A Rikka with Lowbit 线段树

链接：https://www.nowcoder.com/acm/contest/148/A
来源：牛客网

题目描述

Today, Rikka is going to learn how to use BIT to solve some simple data structure tasks. While studying, She finds there is a magic expression $x \& (-x)$ in the template of BIT. After searching for some literature, Rikka realizes it is the implementation of the function $\text{lowbit(x)}$ .

$\text{lowbit}(x)$ is defined on all positive integers. Let a₁...a_m be the binary representation of x while a₁ is the least significant digit, k be the smallest index which satisfies a_k = 1. The value of $\text{lowbit}(x)$ is equal to 2^k-1.

After getting some interesting properties of $\text{lowbit}(x)$ , Rikka sets a simple data structure task for you:

At first, Rikka defines an operator f(x), it takes a non-negative integer x. If x is equal to 0, it will return 0. Otherwise it will return $x-\text{lowbit}(x)$ or $x+\text{lowbit}(x)$ , each with the probability of $\frac{1}{2}$ .

Then, Rikka shows a positive integer array A of length n, and she makes m operations on it.

There are two types of operations:
1. 1 L R, for each index i ∈ [L,R], change A_i to f(A_i).
2. 2 L R, query for the expectation value of $\sum_{i=L}^R A_i$ . (You may assume that each time Rikka calls f, the random variable used by f is independent with others.)

输入描述:

The first line contains a single integer t(1 ≤ t ≤ 3), the number of the testcases.

The first line of each testcase contains two integers n,m(1 ≤ n,m ≤ 10

⁵

). The second line contains n integers A

(1 ≤ A

 ≤ 10

⁸

). 

And then m lines follow, each line contains three integers t,L,R(t ∈ {1,2}, 1 ≤ L ≤ R ≤ n).

输出描述:

For each query, let w be the expectation value of the interval sum, you need to output

$(w \times 2^{nm}) \mod 998244353$

. 

It is easy to find that w x 2

^nm

 must be an integer.

输入例子:

输出例子:

-->

示例1

输入

复制

输出

复制

1572864

1572864

1572864

分析：每次变化+-lowbit得到的期望是原结果，所以我们直接求一下区间和就可以
AC代码：

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn=1e5+10;

const ll mod = 998244353;

ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

ll array1[maxn];

struct node1

{

    ll sum,addmark;

} Node[maxn<<2];

void pushup(int node)

{

    Node[node].sum=Node[node<<1].sum+Node[node<<1|1].sum;

}

void buildtree(int node,int left,int right)

{

    Node[node].addmark=0;

    if(left==right)

    {

        Node[node].sum=array1[left];

        return;

    }

    int m=(left+right)>>1;

    buildtree(node<<1,left,m);

    buildtree(node<<1|1,m+1,right);

    pushup(node);

}

void pushdown(int node,int left,int right)

{

    if(Node[node].addmark)

    {

        int m=(left+right)>>1;

        Node[node<<1].addmark+=Node[node].addmark;

        Node[node<<1|1].addmark+=Node[node].addmark;

        Node[node<<1].sum+=(m-left+1)*Node[node].addmark;

        Node[node<<1|1].sum+=(right-m)*Node[node].addmark;

        Node[node].addmark=0;

    }

}

ll query(int node,int begin,int end,int left,int right)

{

    if(left<=begin&&right>=end)

        return Node[node].sum;

    pushdown(node,begin,end);

    int m=(begin+end)>>1;//小心!!

    ll ans=0;

    if(left<=m)

        ans+=query(node<<1,begin,m,left,right);

    if(right>m)

        ans+=query(node<<1|1,m+1,end,left,right);

    return ans;

}

void update(int node,int add,int begin,int end,int left,int right)

{

    if(begin>=left&&end<=right)

    {

        Node[node].sum+=(end-begin+1)*add;

        Node[node].addmark+=add;

        return;

    }

    int m=(begin+end)>>1;//小心!!

    pushdown(node,begin,end);

    if(left<=m)

        update(node<<1,add,begin,m,left,right);

    if(right>m)

        update(node<<1|1,add,m+1,end,left,right);

    pushup(node);

}

ll n,m;

int main()

{

    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

    int i,j,k,t;

    cin>>t;

    while(t--)

    {

        cin>>n>>m;

        for(i=1; i<=n; i++) cin>>array1[i];

        buildtree(1,1,n);

        for(i=1; i<=m; i++)

        {

            int a,b,c;

            cin>>a>>b>>c;

            if(a==2) cout<<query(1,1,n,b,c)%mod*qp(2,n*m)%mod<<endl;

        }

    }

    return 0;

}