清神的线段树方法二_暂存

原创于 2019-05-04 23:04:05 发布 · 174 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#线段树

暂存代码专栏收录该内容

2 篇文章

订阅专栏

博客围绕线段树展开，但具体内容缺失。线段树是信息技术领域的数据结构，在算法和数据处理中有重要应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#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))

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
}*/
#define mid ((l+r)>>1)
const int MAX_N = 1025;
int maxx[MAX_N<<2],maxx_[MAX_N<<2],minn[MAX_N<<2],arr[MAX_N];
void up(int rt)
{
    minn[rt] = min(minn[rt<<1],minn[rt<<1|1]);
    maxx[rt] = max(maxx[rt<<1],maxx[rt<<1|1]);
    maxx_[rt] = max(max(maxx_[rt<<1],maxx_[rt<<1|1]),max(maxx[rt<<1]*maxx[rt<<1|1],minn[rt<<1]*minn[rt<<1|1]));
}
void build(int rt,int l,int r)
{
    minn[rt] = inf,maxx[rt] =maxx_[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);
}
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)
{
    if(x<=l&&r<=y)
    {
        return maxx_[rt];
    }
    if(x>mid) return query(rt<<1|1,mid+1,r,x,y);
    else if(y<=mid) return query(rt<<1,l,mid,x,y);
    else return max(max(query(rt<<1,l,mid,x,y),query(rt<<1|1,mid+1,r,x,y)),max(Query(rt<<1,l,mid,x,y,1)*Query(rt<<1|1,mid+1,r,x,y,1),Query(rt<<1,l,mid,x,y,0)*Query(rt<<1|1,mid+1,r,x,y,0)));
}
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--)
        {
            scanf("%d%d",&x,&y);
            printf("%d\n",query(1,1,n,x,y));
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}