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;
}