萌新线段树求调阿巴阿巴

P3372 【模板】线段树 1

Wunsch @ 2023-01-26 18:14:03

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#define ll long long
#define LS o<<1
#define RS LS|1
using namespace std;

ll n,m,t[500001],a[500001],laz[500001],sum,x,y,k,q;

void build(ll o,ll l,ll r){
    if(l==r){cin>>t[o];return;}
    ll mid=l+r>>1;
    build(LS,l,mid);
    build(RS,mid+1,r);
    t[o]=t[LS]+t[RS];
}

void down(ll o,ll l,ll r){
    if(l==r||!laz[o])return;
    ll mid=l+r>>1;
    laz[LS]+=laz[o];
    laz[RS]+=laz[o];
    t[LS]+=laz[o]*(mid-l+1);
    t[RS]+=laz[o]*(r-mid);
    laz[o]=0;
}

void change(ll o,ll l,ll r,ll x,ll y,ll sum){
    if(x<=l&&r<=y){laz[o]+=sum;t[o]+=(r-l+1)*sum;return;}
    down(o,l,r);
    ll mid=l+r>>1;
    if(x<=mid)change(LS,l,mid,x,y,sum);
    if(y>mid)change(RS,mid+1,r,x,y,sum);
    t[o]=t[LS]+t[RS];
}

ll qury(ll o,ll l,ll r,ll x,ll y){
    if(x<=l&&r<=y)return t[o];
    down(o,l,r);
    ll ret=0;
    ll mid=l+r>>1;
    if(x<=mid)ret+=qury(LS,l,mid,x,y);
    if(y>mid)ret+=qury(RS,mid+1,r,x,y);
    return ret;
}

int main(){
    cin>>n>>m;

    build(1,1,n);
    for(int i=1;i<=m;i++){
        cin>>q;
        if(q==1){
            cin>>x>>y>>k;change(1,1,n,x,y,k);
        }
        else{
            cin>>x>>y;cout<<qury(1,1,n,x,y);
        }
    }
    return 0;
}

by Eleveslaine @ 2023-01-26 18:17:43

头一次见写线段树不用结构体的。
您不存每个区间的 l,r


by Coffins @ 2023-01-26 18:20:39

@Franz_Liszt 不用结构体完全可以啊,我就没用过结构体


by Acoipp @ 2023-01-26 18:21:21

换行呢?


by Eleveslaine @ 2023-01-26 18:21:49

@Coffins 好吧,但是个人认为结构体稍微好写点


by Acoipp @ 2023-01-26 18:21:50

cout<<qury(1,1,n,x,y)<<endl;

by hello_world_djh @ 2023-01-26 18:22:52

@Wunsch 忘了换行了


by Coffins @ 2023-01-26 18:23:35

@Wunsch 差个换行就能A了


by Xy_top @ 2023-01-26 18:38:02

@Wunsch 不建议用 endl 换行,非常非常非常慢,可以用 "\n" 换行,快很多


by Wunsch @ 2023-01-26 18:44:26

@hello_world_djh (好吧我是傻子 不过谢谢


by Wunsch @ 2023-01-26 18:44:56

@Coffins okok谢谢


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