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 in the template of BIT. After searching for some literature, Rikka realizes it is the implementation of the function .

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 is equal to 2^k-1.

After getting some interesting properties of , 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 or , each with the probability of .

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 . (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

_i

(1 ≤ A

_i

 ≤ 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 

. 

It is easy to find that w x 2

^nm

 must be an integer.

示例1

输入

复制

1
3 6
1 2 3
1 3 3
2 1 3
1 3 3
2 1 3
1 1 3
2 1 3

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;
}