luogu P1438 无聊的数列

思路

区间\(L\)到\(R\)内加等差数列
已知首项为\(K\)，公差为\(D\)
那么每一位加的数值为
\[K+(i-L)*D(L<=i<=R)\]
\[K+i*D-L*D(L<=i<=R)\]
\[K-L*D+i*D(L<=i<=R)\]
我们可以分别加一下\(K-L*D\)和\(i*D\)
前者可以算出来，直接区间加就可以
后者的话，先维护一下\([L,R]\)内的下标和
结合律用一下，加上R*下标和就行了

update

修改的时候在区间\([L,R]\)内的贡献为
\[(K-L*D)*(R-L+1)+\sum_{L}^{R}*D\]

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define ls rt<<1
#define rs rt<<1|1
#define ll long long
#define FOR(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
const int maxn = 1e5 + 7;
int read() {
    int x = 0, f = 1; char s = getchar();
    for (; s < '0' || s > '9'; s = getchar()) if (s == '-') f = -1;
    for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;
}
struct node {
    int l,r,size;
    ll ans,lazy1,tot,lazy2;
}e[maxn<<2];
int n,m,a[maxn];
void pushup(int rt) {
    e[rt].ans=e[ls].ans+e[rs].ans;
}
void pushdown(int rt) {
    if(e[rt].lazy1) {
        e[ls].ans+=e[ls].size*e[rt].lazy1;
        e[rs].ans+=e[rs].size*e[rt].lazy1;
        e[ls].lazy1+=e[rt].lazy1;
        e[rs].lazy1+=e[rt].lazy1;
        e[rt].lazy1=0;
    }
    if(e[rt].lazy2) {
        e[ls].ans+=e[ls].tot*e[rt].lazy2;
        e[rs].ans+=e[rs].tot*e[rt].lazy2;
        e[ls].lazy2+=e[rt].lazy2;
        e[rs].lazy2+=e[rt].lazy2;
        e[rt].lazy2=0;
    }
}
void build(int l,int r,int rt) {
    e[rt].l=l,e[rt].r=r,e[rt].size=r-l+1;
    if(l==r) {
        e[rt].tot=l;
        e[rt].ans=a[l];
        return;
    }
    int mid=(e[rt].l+e[rt].r)>>1;
    build(l,mid,ls);
    build(mid+1,r,rs);
    pushup(rt);
    e[rt].tot=e[ls].tot+e[rs].tot;
}
void modfity_1(int L,int R,ll k,int rt) { // 区间加
    if(L<=e[rt].l&&e[rt].r<=R) {
        e[rt].ans+=k*e[rt].size;
        e[rt].lazy1+=k; 
        return;
    }
    pushdown(rt);
    int mid=(e[rt].l+e[rt].r)>>1;
    if(L<=mid) modfity_1(L,R,k,ls);
    if(R>mid) modfity_1(L,R,k,rs);
    pushup(rt);
}
void modfity_2(int L,int R,ll k,int rt) { // 区间i*k
    if(L<=e[rt].l&&e[rt].r<=R) {
        e[rt].ans+=k*e[rt].tot;
        e[rt].lazy2+=k;
        return;
    }
    pushdown(rt);
    int mid=(e[rt].l+e[rt].r)>>1;
    if(L<=mid) modfity_2(L,R,k,ls);
    if(R>mid) modfity_2(L,R,k,rs);
    pushup(rt); 
}
ll query(int L,int rt) {
    if(e[rt].l==e[rt].r) {
        return e[rt].ans;
    }
    pushdown(rt);
    int mid=(e[rt].l+e[rt].r)>>1;
    if(L<=mid) return query(L,ls);
    else return query(L,rs);
}
int main() {
    n=read(),m=read();
    FOR(i,1,n) a[i]=read();
    build(1,n,1);
    FOR(i,1,m) {
        int pd=read();
        if(pd==1) {
            int l=read(),r=read(),k=read(),d=read();
            modfity_1(l,r,k-l*d,1);
            modfity_2(l,r,d,1);
        } else {
            int x=read();
            printf("%lld\n",query(x,1));
        }
    }
    return 0;
}