solution

用线段树维护即可。
考虑线段树的懒标记如何维护。我们维护两个懒标记添加到当前位置的数列在该区间开始位置的首项之和,表示这些添加的数列公差之和。

因为如果往一个位置添加了一个等差数列。其中某个位置可以看作,那么再添加一个等差数列,那么该位置就可以看作,相当于添加了一个首项为公差为的等差数列。

挺简单的思路,就是有点难调233

code

/*
* @Author: wxyww
* @Date:   2020-04-17 19:25:25
* @Last Modified time: 2020-04-17 20:49:01
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 200010,mod = 223092870;
#define int ll
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1; c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0'; c = getchar();
    }
    return x * f;
}
int tree[N << 2],lazy1[N << 2],lazy2[N << 2];

int a[N];

void build(int rt,int l,int r) {
    if(l == r) {
        tree[rt] = a[l];return;
    }
    int mid = (l + r) >> 1;
    build(rt << 1,l,mid);
    build(rt << 1 | 1,mid + 1,r);
    tree[rt] = (tree[rt << 1] + tree[rt << 1 | 1]) % mod;
}
void upd(int &x,int y) {
    x = (x + y) % mod;
}
void pushdown(int rt,int ln,int rn,int l,int r) {
    if(lazy1[rt] || lazy2[rt]) {
        upd(lazy1[rt << 1],lazy1[rt]);
        upd(lazy2[rt << 1],lazy2[rt]);
        upd(lazy2[rt << 1 | 1],lazy2[rt]);

        ll f1 = lazy1[rt],f2 = (lazy1[rt] + 1ll * ln * lazy2[rt]) % mod;

        upd(lazy1[rt << 1 | 1],f2);

        upd(tree[rt << 1],1ll * (f1 + f1 + (ln - 1) * lazy2[rt]) * ln / 2 % mod);
        upd(tree[rt << 1 | 1],1ll * (f2 + f2 +  1ll * (rn - 1) * (lazy2[rt])) * rn / 2 % mod);
        lazy1[rt] = lazy2[rt] = 0;
    }
}

ll query(int rt,int l,int r,int L,int R) {
    if(L <= l && R >= r) return tree[rt];
    int mid = (l + r) >> 1;

    pushdown(rt,mid - l + 1,r - mid,l,r);
    int ret = 0;
    if(L <= mid) ret += query(rt << 1,l,mid,L,R);
    if(R > mid) ret += query(rt << 1 | 1,mid + 1,r,L,R);
    return ret % mod;
}
void update(int rt,int l,int r,int L,int R,ll x,ll d) {
    ll xx = x + (l - L) * d;
    // printf("%lld %lld %lld %lld\n",l,r,x,xx);
    if(L <= l && R >= r) {
        // printf("%lld %lld\n",l,r);
        tree[rt] += 1ll * (xx + xx + 1ll * d * (r - l)) * (r - l + 1) / 2 % mod;
        tree[rt] %= mod;
        // printf("%lld %lld %lld\n",rt,l,r);
        // printf("%lld %lld %lld\n",l,r,(xx + xx + 1ll * d * (r - l)) * (r - l + 1) / 2);

        lazy1[rt] += xx;lazy2[rt] += d;
        lazy1[rt] %= mod;lazy2[rt] %= mod;
        return;
    }


    int mid = (l + r) >> 1;
    pushdown(rt,mid - l + 1,r - mid,l,r);
    if(L <= mid) update(rt << 1,l,mid,L,R,x,d);
    if(R > mid) update(rt << 1 | 1,mid + 1,r,L,R,x,d);
    tree[rt] = (tree[rt << 1] + tree[rt << 1 | 1]) % mod;
}


signed main() {
    // freopen("1.in","r",stdin);
    int n = read();
    for(int i = 1;i <= n;++i) a[i] = read();
    build(1,1,n);
    int Q = read();
    while(Q--) {
        int opt = read();
        if(opt == 1) {
            int l = read(),r = read(),x = read() % mod,d = read() % mod;
            update(1,1,n,l,r,x,d);
        }
        else {
            int l = read(),r = read(),m = read();
            printf("%lld\n",query(1,1,n,l,r) % m);
        }
        // puts("!!!");
        // printf("!!%lld\n",lazy1[2]);
    }
    return 0;
}