清神的线段树方法一_暂存

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 1025;
int maxx[MAX_N<<2],minn[MAX_N<<2],arr[MAX_N];
int Max(int a,int b){if(a>b) return a;return b;}
int Min(int a,int b){if(a<b) return a;return b;}
void up(int rt)
{
    maxx[rt] = Max(maxx[rt<<1],maxx[rt<<1|1]);
    minn[rt] = Min(minn[rt<<1],minn[rt<<1|1]);
}
void build(int rt,int l,int r)
{
    maxx[rt] = -inf;minn[rt] = inf;
    if(l==r)
    {
        maxx[rt] = minn[rt] = arr[l];
        return ;
    }
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    up(rt);
}
void update(int rt,int l,int r,int x,int v,int v_)
{
    if(l==r)
    {
        maxx[rt] = v;
        minn[rt] = v_;
        return ;
    }
    if(x<=mid) update(rt<<1,l,mid,x,v,v_);
    else update(rt<<1|1,mid+1,r,x,v,v_);
    up(rt);
}
int Query(int rt,int l,int r,int x,int y,int v)// v = 1 最大值 v = 0 最小值
{
    if(x<=l&&r<=y)
    {
        if(v==1) return maxx[rt];
        return minn[rt];
    }
    if(x>mid)   return Query(rt<<1|1,mid+1,r,x,y,v);
    else if(y<=mid) return Query(rt<<1,l,mid,x,y,v);
    else
    {
        if(v==1) return Max(Query(rt<<1,l,mid,x,y,v),Query(rt<<1|1,mid+1,r,x,y,v));
        return Min(Query(rt<<1,l,mid,x,y,v),Query(rt<<1|1,mid+1,r,x,y,v));
    }
}
int query(int rt,int l,int r,int x,int y,int v)//找最大值的下标
{
    if(maxx[rt]<v) return -1;
    if(l==r)
    {
        return l;
    }
    int ans = -1;
    if(x<=mid&&maxx[rt<<1]>=v) ans = query(rt<<1,l,mid,x,y,v);
    if(ans==-1) ans = query(rt<<1|1,mid+1,r,x,y,v);
    return ans;
}
int query_(int rt,int l,int r,int x,int y,int v)//找最小值的下标
{
    if(minn[rt]>v) return -1;
    if(l==r)
    {
        return l;
    }
    int ans = -1;
    if(x<=mid&&minn[rt<<1]<=v) ans = query_(rt<<1,l,mid,x,y,v);
    if(ans==-1) ans = query_(rt<<1|1,mid+1,r,x,y,v);
    return ans;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int t,n,Q,x,y;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
        build(1,1,n);
        scanf("%d",&Q);
        while(Q--)
        {
            int ans = -inf;
            scanf("%d%d",&x,&y);
            int maxx_ = Query(1,1,n,x,y,1),minn_;
            int xb = query(1,1,n,x,y,maxx_);
            update(1,1,n,xb,-inf,-inf);
            ans = max(ans,maxx_*Query(1,1,n,x,y,1));
            update(1,1,n,xb,maxx_,maxx_);
            minn_ = Query(1,1,n,x,y,0);
            xb = query_(1,1,n,x,y,minn_);
            update(1,1,n,xb,inf,inf);
            ans = max(ans,minn_*Query(1,1,n,x,y,0));
            update(1,1,n,xb,minn_,minn_);
            printf("%d\n",ans);
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}