[hiho1586]Minimum

题意：区间内乘积最小值，带修改。
解题关键：线段树裸题，考场上以为x y必须不等，还维护了次小值，尼玛嗨尼玛嗨，划水一整场，心态爆炸。

注意坐标需要+1

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<iostream>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=131073;
ll a[maxn],tree1[maxn<<2],tree2[maxn<<2];
void pushup(int rt){
    tree1[rt]=max(tree1[rt<<1|1],tree1[rt<<1]);
    tree2[rt]=min(tree2[rt<<1|1],tree2[rt<<1]);
}
void build(ll rt,ll l,ll r){
    if(l==r){
        tree1[rt]=tree2[rt]=a[l];
        return;
    }
    ll mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    pushup(rt);
}

void update(ll rt,ll l,ll r,ll p,ll c){
    if(l>r) return;
    if(l==r){
        tree1[rt]=tree2[rt]=c;
        return;
    }
    ll mid=(l+r)>>1;
    if(p<=mid) update(rt<<1, l, mid, p, c);
    else update(rt<<1|1, mid+1, r, p, c);
    pushup(rt);
}

ll query1(ll rt,ll l,ll r,ll tl,ll tr){
    if(tl>tr) return -inf;
    if(tl<=l&&tr>=r){
        return tree1[rt];
    }
    ll mid=(l+r)>>1,res=-inf;
    if(tl<=mid) res=max(res,query1(rt<<1,l,mid,tl,tr));
    if(tr>mid) res=max(query1(rt<<1|1,mid+1,r,tl,tr),res);
    return res;
}

ll query2(ll rt,ll l,ll r,ll tl,ll tr){
    if(tl>tr) return inf;
    if(tl<=l&&tr>=r){
        return tree2[rt];
    }
    ll mid=(l+r)>>1,res=inf;
    if(tl<=mid) res=min(res,query2(rt<<1,l,mid,tl,tr));
    if(tr>mid) res=min(query2(rt<<1|1,mid+1,r,tl,tr),res);
    return res;
}

int main(){
    ll n,m;
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin>>t;
    while(t--){
        cin>>n;
        n=1<<n;
        memset(tree1, 0, sizeof tree1);
        memset(tree2,0,sizeof tree2);
        memset(a,0,sizeof a);
        for(int i=1;i<=n;i++) cin>>a[i];
        build(1, 1, n);
        cin>>m;
        for(int i=0;i<m;i++){
            int a,b,c;
            cin>>a>>b>>c;
            if(a==1){
                ll ans1=query1(1,1,n,b+1,c+1);
                ll ans2=query2(1,1,n,b+1,c+1);
                if(ans1>=0&&ans2<0){
                    cout<<1ll*ans1*ans2<<"
";
                    continue;
                }
                else if(ans1>=0&&ans2>=0){
                    cout<<1ll*ans2*ans2<<"
";
                    continue;
                }
                else{
                    cout<<1ll*ans1*ans1<<"
";
                }
            }
            else{
                update(1, 1, n, b+1, c);
            }
        }
    }

}