【HDU 5828 Rikka with Sequence】线段树 区间平方 区间加 区间求和

HDU 5828
题意 给你n个数 有三个操作 1 l r v 给 l - r的数加上v
2 l r 给 l - r 的数都开方
3 l r 求 l - r的数和
从最简单的情况开始 如果没有1操作 只有区间开方 和区间求和 可以参考这题博客
BZOJ 3211
问题是有相加
我们要知道开方有个性质 次数开多了以后 这个区间的maxx - minn的值是会越来越小的
就是这个公式
$$\sqrt{x+d}-\sqrt{y+d}<=\sqrt{x}-\sqrt{y}$$
那么我们知道 当开方的时候 如果这个区间maxx <= 1 那么我们就不用进行开方操作
如果这个区间 maxx - minn > 1 我们则要进行开方操作
为什么在maxx-minn等于1的时候需要特判呢
因为我们发现 例如 3 4 这两个数 他们原先的maxx 是 4 minn 是 3
开过方以后 maxx 是 2 minn 是 1
你会发现 4 -> 2 , 3 -> 1这不是区间加的过程么 所以这可以用区间加优化
还有一种情况 2 3 maxx 是 3 minn 是 2
开过方以后 maxx = minn = 1
那不是区间替换的过程吗？
就是利用这个性质达到优化时间的操作
所以 2 操作就变成了 要吗不操作
要吗区间加
要吗区间特换
要吗暴力修改
然后一开始s没注意全是long long WA 了两发

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#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 lc (rt<<1)
#define rc (rt<<11)
#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);
const int MAX_N = 100025;
int maxx[MAX_N<<2],minn[MAX_N<<2],col[MAX_N<<2],seto[MAX_N<<2];
ll s[MAX_N<<2];
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)
{
    s[rt] = s[rt<<1] + s[rt<<1|1];
    maxx[rt] = Max(maxx[rt<<1],maxx[rt<<1|1]);
    minn[rt] = Min(minn[rt<<1],minn[rt<<1|1]);
}
void down(int rt,int l,int r)
{
    if(seto[rt]!=-1)
    {
        s[rt<<1] = 1ll*(mid-l+1)*seto[rt];
        s[rt<<1|1] = 1ll*(r-mid)*seto[rt];
        maxx[rt<<1] = maxx[rt<<1|1] = seto[rt];
        minn[rt<<1] = minn[rt<<1|1] = seto[rt];
        seto[rt<<1] = seto[rt<<1|1] = seto[rt];
        col[rt<<1] = col[rt<<1|1] = 0;
        seto[rt]=-1;
    }
    if(col[rt])
    {
        s[rt<<1] += 1ll*(mid-l+1)*col[rt];
        s[rt<<1|1] += 1ll*(r-mid)*col[rt];
        col[rt<<1] += col[rt];
        col[rt<<1|1] += col[rt];
        maxx[rt<<1] += col[rt];
        maxx[rt<<1|1] += col[rt];
        minn[rt<<1] += col[rt];
        minn[rt<<1|1] += col[rt];
        col[rt] = 0;
    }
}
void build(int rt,int l,int r)
{
    seto[rt] = -1;col[rt] = 0;maxx[rt] = minn[rt] = s[rt] = 0;
    if(l==r)
    {
        scanf("%lld",&s[rt]);
        maxx[rt] = minn[rt] = s[rt];
        return ;
    }
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    up(rt);
}
void change(int rt,int l,int r,int x,int y,int v)
{
    if(x<=l&&r<=y)
    {
        s[rt] += 1ll*v*(r-l+1);
        col[rt] += v;
        maxx[rt] += v;
        minn[rt] += v;
        return ;
    }
    down(rt,l,r);
    if(x<=mid) change(rt<<1,l,mid,x,y,v);
    if(mid<y) change(rt<<1|1,mid+1,r,x,y,v);
    up(rt);
}
void update(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
    {
        if(maxx[rt]<=1) return;
        if(l==r)
        {
            s[rt] = floor(sqrt(s[rt]));
            maxx[rt] = minn[rt] = s[rt];
            return ;
        }
        if(maxx[rt] == minn[rt])
        {
            int tmp = maxx[rt];
            maxx[rt] = minn[rt] = floor(sqrt(tmp));
            seto[rt] = maxx[rt];col[rt] = 0;
            s[rt] = 1ll*(r-l+1) * maxx[rt];
            return;
        }
        else if(maxx[rt]-minn[rt]==1)
        {
            int ta = maxx[rt],tb = minn[rt];
            maxx[rt] = floor(sqrt(maxx[rt]));
            minn[rt] = floor(sqrt(minn[rt]));
            if(ta-maxx[rt]==tb-minn[rt])
            {
                col[rt] += maxx[rt] - ta;
                s[rt] += 1ll*(maxx[rt]-ta)*(r-l+1);
            }
            else
            {
                col[rt] = 0;seto[rt] = maxx[rt];
                s[rt] = 1ll*(r-l+1)*maxx[rt];
            }
            return;
        }
        down(rt,l,r);
        update(rt<<1,l,mid,x,y);
        update(rt<<1|1,mid+1,r,x,y);
        up(rt);
        return ;
    }
    down(rt,l,r);
    if(x<=mid) update(rt<<1,l,mid,x,y);
    if(mid<y) update(rt<<1|1,mid+1,r,x,y);
    up(rt);
}
ll query(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
    {
        return s[rt];
    }
    ll res =0;
    down(rt,l,r);
    if(x<=mid) res+= query(rt<<1,l,mid,x,y);
    if(mid<y) res+=query(rt<<1|1,mid+1,r,x,y);
    return res;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int t,n,Q,opt,x,y,v;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&Q);
        build(1,1,n);
        while(Q--)
        {
            scanf("%d%d%d",&opt,&x,&y);
            if(opt==1)
            {
                scanf("%d",&v);
                change(1,1,n,x,y,v);
            }
            else if(opt==2)
            {
                update(1,1,n,x,y);
            }
            else
            {
                printf("%lld\n",query(1,1,n,x,y));
            }
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}