鏈接：https://ac.nowcoder.com/acm/contest/148/A
來源：牛客網(wǎng)

Rikka with Lowbit
時間限制：C/C++ 5秒，其他語言10秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
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)x&(?x) in the template of BIT. After searching for some literature, Rikka realizes it is the implementation of the function \text{lowbit(x)}lowbit(x).

\text{lowbit}(x)lowbit(x) 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 \text{lowbit}(x)lowbit(x) is equal to 2k-1.

After getting some interesting properties of \text{lowbit}(x)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)x?lowbit(x) or x+\text{lowbit}(x)x+lowbit(x), each with the probability of \frac{1}{2}
2
1
?
.

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 L R, for each index i ∈ [L,R], change Ai to f(Ai).

2 L R, query for the expectation value of \sum_{i=L}^R A_i∑
i=L
R
?
A
i
?
. (You may assume that each time Rikka calls f, the random variable used by f is independent with others.)
輸入描述:
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).
輸出描述:
For each query, let w be the expectation value of the interval sum, you need to output (w \times 2^{nm}) \mod 998244353(w×2
nm
)mod998244353.

It is easy to find that w x 2nm must be an integer.
示例1
輸入
復(fù)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
輸出
復(fù)制
1572864
1572864
1572864

題意：

思路：

分析會發(fā)現(xiàn)操作1對數(shù)字的期望根本就不會改變，于是就是一個裸的區(qū)間查詢（用前綴和都可以做）（記得取模）。

細(xì)節(jié)見代碼：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=100010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ll tree[maxn];ll a[maxn];int n,m; int lowbit(int x) {return x&(-x); } const ll mod=998244353ll; void add(int id,ll x) {while(id<=n){tree[id]=(tree[id]+x)%mod;id+=lowbit(id);} } ll ask(int id) {ll res=0ll;while(id){res=(res+tree[id])%mod;id-=lowbit(id);}return res; } int main() {//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);gbtb;int t;cin>>t;while(t--){MS0(tree);cin>>n>>m;repd(i,1,n){cin>>a[i];add(i,a[i]);}ll ans;int op,l,r;repd(i,1,m){cin>>op>>l>>r;if(op==1){continue;}ans=(ask(r)-ask(l-1)+mod)%mod;ans=(ans*powmod(2ll,1ll*n*m,mod))%mod;cout<<ans<<endl;}}return 0; }inline void getInt(int* p) {char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}}else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 + ch - '0';}} }

轉(zhuǎn)載于:https://www.cnblogs.com/qieqiemin/p/11261667.html

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