<JZOJ5943>树

一开始t了五个点我就一脸懵逼

然后 发现高级操作...

就是那个tor的数组2333

可以让一些不需要改的不再去改啦

位运算果然是神奇的东西XD

魔性哈哈哈

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define rint register int
#define MOD 998244353
template <class T>inline void read(T &X)
{
    X=;int W=;char ch=;
    while(!isdigit(ch))W|=ch=='-',ch=getchar();
    while(isdigit(ch))X=(X<<)+(X<<)+(ch^),ch=getchar();
    X=W?-X:X;return;
}

int n,Q,k;
long long ans(),ans2(),tree[],tpow[],tor[],a[];
void build(int l,int r,int pos)
{
    if(l==r)
    {
        tree[pos]=a[l];
        tpow[pos]=(a[l]*a[l])%MOD;
        tor[pos]=a[l];
        return;
    }
    int mid=l+r>>;
    build(l,mid,pos<<);
    build(mid+,r,pos<<|);

    tree[pos]=tree[pos<<]+tree[pos<<|];
    tpow[pos]=(tpow[pos<<]+tpow[pos<<|])%MOD;
    tor[pos]=(tor[pos<<]|tor[pos<<|]);
}

void query(int l,int r,int L,int R,int pos)
{
    if(L<=l && r<=R)
    {
        ans+=tree[pos];
        ans2=(ans2+tpow[pos])%MOD;
        return;
    }
    int mid=l+r>>;
    if(L<=mid)query(l,mid,L,R,pos<<);
    if(R>mid)query(mid+,r,L,R,pos<<|);
}

void change(int l,int r,int L,int R,int pos)
{
    if((tor[pos]&(~k))==)return;
    if(l==r)
    {
        tree[pos]=tree[pos]&k;
        a[l]=a[l]&k;
        tor[pos]=a[l];
        tpow[pos]=(a[l]*a[l])%MOD;
        return;
    }
    int mid=l+r>>;
    if(L<=mid)change(l,mid,L,R,pos<<);
    if(R>mid)change(mid+,r,L,R,pos<<|);

    tree[pos]=tree[pos<<]+tree[pos<<|];
    tpow[pos]=(tpow[pos<<]+tpow[pos<<|])%MOD;
    tor[pos]=(tor[pos<<]|tor[pos<<|]);
}

int main()
{
//    freopen("seg.in","r",stdin);
//    freopen("seg.out","w",stdout);
    read(n);
    for(rint i=;i<=n;++i)read(a[i]);
    build(,n,);
    read(Q);
    while(Q--)
    {
        int num,l,r;
        read(num);
        if(num==)
        {
            read(l),read(r),read(k);
            change(,n,l,r,);
        continue;
        }
        else
        {
            read(l),read(r);
            ans=;ans2=;
            query(,n,l,r,);
            if(num==)
            {
                printf("%lld\n",ans);
            continue;
            }
            if(num==)
            {
                ans%=MOD;
                ans=(*ans2%MOD*(r-l+)%MOD+*ans%MOD*ans)%MOD;
                printf("%lld\n",ans);
            continue;
            }
        }
    }
}