【模板】线段树区间加、乘

只需要使用两个lazy-tag(addv, mulv)来维护，一个维护区间加，一个维护区间乘。

注意mulv标记应在建树的时候初始化成1.

在下传标记的时候，mulv下传时会对addv造成影响，所以修改addv时要考虑到mulv

一定要开long long，也要检查Update, down这些函数里需要开long long的有没有开

【代码：】

//线段树维护区间加、乘
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int MAXN = 500000 + 1;
const int ADD = 1, MUL = 2;

int n, m, p;
long long a[MAXN];
struct Segment {
    long long sum, addv, mulv;
}t[MAXN << 2];

inline long long read() {
    long long x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - 48, ch = getchar();
    return x * f;
}

void Build(int o, int l, int r) {
    t[o].mulv = 1;
    if(l == r) t[o].sum = a[l];
    else {
        int mid = (l + r) >> 1;
        Build(o << 1, l, mid);
        Build(o << 1|1, mid + 1, r);
        t[o].sum = (t[o << 1].sum + t[o << 1|1].sum) % p;
    }
}

inline void down(int o, int len) {
    long long mul = t[o].mulv, add = t[o].addv;
    t[o << 1].sum = (t[o << 1].sum * mul + add * (len - (len >> 1))) % p;
    t[o << 1|1].sum = (t[o << 1|1].sum * mul + add * (len >> 1)) % p;
    t[o << 1].mulv = (t[o << 1].mulv * mul) % p;
    t[o << 1|1].mulv = (t[o << 1|1].mulv * mul) % p;
    t[o << 1].addv = (t[o << 1].addv * mul + add) % p;
    t[o << 1|1].addv = (t[o << 1|1].addv * mul + add) % p;
    t[o].mulv = 1, t[o].addv = 0;
}
void Update(int o, int l, int r, int ul, int ur, long long v, int fl) {
    if(ul <= l && r <= ur) {
        if(fl == 1) {
            t[o].sum = (t[o].sum + v * (r - l + 1)) % p;
            t[o].addv = (t[o].addv + v) % p;
        }
        else if(fl == 2) {
            t[o].sum = (t[o].sum * v) % p;
            t[o].mulv = (t[o].mulv * v) % p;
            t[o].addv = (t[o].addv * v) % p;
        }
    }
    else {
        if(t[o].mulv != 1 || t[o].addv)
            down(o, r - l + 1);
        int mid = (l + r) >> 1;
        if(ul <= mid) Update(o << 1, l, mid, ul, ur, v, fl);
        if(ur > mid) Update(o << 1|1, mid + 1, r, ul, ur, v, fl);
        t[o].sum = (t[o << 1].sum + t[o << 1|1].sum) % p;
    }
}

long long Query(int o, int l, int r, int ql, int qr) {
    if(ql <= l && r <= qr) return t[o].sum % p;
    int mid = (l + r) >> 1;
    long long ret = 0;
    if(t[o].mulv != 1 || t[o].addv)
        down(o, r - l + 1);
    if(ql <= mid) ret += Query(o << 1, l, mid, ql, qr);
    if(qr > mid) ret += Query(o << 1|1, mid + 1, r, ql, qr);
    return ret % p;
}

int main() {
    n = read(), m = read(), p = read();
    for(int i=1; i<=n; ++i)
        a[i] = read();
    Build(1, 1, n);
    while(m--) {
        long long fl = read(), x = read(), y = read();
        if(fl == 1) {
            long long k = read();
            Update(1, 1, n, x, y, k, MUL);
        }
        else if(fl == 2) {
            long long k = read();
            Update(1, 1, n, x, y, k, ADD);
        }
        else printf("%lld
", Query(1, 1, n, x, y));
    }
}