牛客第十场A

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...am be the binary representation of x while a1 is the least significant digit, k be the smallest index which satisfies ak = 1. The value of  is equal to 2k-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 Ai to f(Ai).
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 ≤ 105). The second line contains n integers Ai(1 ≤ Ai ≤ 108). 

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 2nm must be an integer.

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

加和减概率相同，加减的数也相同所以不变

函数f(x)它的值有一半概率是x - lowbit(x)，一半概率是x + lowbit(x)

树状数组

#include<bits/stdc++.h>
using namespace std;
#define mod 998244353
long long  wwf(long long  a, long long b)
{
    long long ans = 1;
    a %= mod;
    while (b) {
        if (b & 1)ans = ans * a%mod;
        a = a * a%mod;
        b >>= 1;
    }
    return ans;
}
int main()
{
    long long t,n,m,b,c,l,r,d,a[100005];
    long long ans=0,w,we;
    cin>>t;
    while(t--) {
        scanf("%lld %lld",&n,&m);
        ans=wwf(2,n*m);
        a[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%lld",&c);
            a[i]=a[i-1]+c;
        }
        for(int i=0;i<m;i++){
            scanf("%lld %lld %lld",&d,&l,&r);
            if(d==2){
                we=a[r]-a[l-1]+mod;
                we%=mod;
                printf("%lld\n",we*ans%mod);
            }
        }
 
    }
}