P3373 线段树2（多重标记线段树）题解

题意：

操作有：区间加，区间乘，区间询问求和

思路：

设一个数为(m*sum+a)，加就变成了(m*sum+a+a_2)，乘就变成了(m*m_2*sum+a*m_2)，所以我们设两个标记(mul)表示乘，(add)表示加，然后如上转化。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100000 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 10007;
using namespace std;
ll n, m, p;
ll sum[maxn << 2], add[maxn << 2], mul[maxn << 2];
int a[maxn];
void pushup(int rt){
    sum[rt] = (sum[rt << 1] + sum[rt << 1 | 1]) % p;
}
void pushdown(int rt, int l, int r){
    int m = (l + r) >> 1;
    sum[rt << 1] = (sum[rt << 1] * mul[rt] + add[rt] * (m - l + 1)) % p;
    sum[rt << 1 | 1] = (sum[rt << 1 | 1] * mul[rt] + add[rt] * (r - m)) % p;
    add[rt << 1] = (add[rt << 1] * mul[rt] + add[rt]) % p;
    add[rt << 1 | 1] = (add[rt << 1 | 1] * mul[rt] + add[rt]) % p;
    mul[rt << 1] = mul[rt << 1] * mul[rt] % p;
    mul[rt << 1 | 1] = mul[rt << 1 | 1] * mul[rt] % p;
    add[rt] = 0;
    mul[rt] = 1;
}
void build(int l, int r, int rt){
    add[rt] = 0;
    mul[rt] = 1;
    if(l == r){
        sum[rt] = a[l];
        return;
    }
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    pushup(rt);
}
void update(int L, int R, int l, int r, int v, int op, int rt){
    if(L <= l && R >= r){
        if(op == 2){
           add[rt] = (add[rt] + v) % p;
           sum[rt] = (sum[rt] + (r - l + 1) * v) % p;
        }
        else{
           add[rt] = add[rt] * v % p;
           mul[rt] = mul[rt] * v % p;
           sum[rt] = sum[rt] * v % p;
        }
        return;
    }
    pushdown(rt, l, r);
    int m = (l + r) >> 1;
    if(L <= m)
        update(L, R, l, m, v, op, rt << 1);
    if(R > m)
        update(L, R, m + 1, r, v, op, rt << 1 | 1);
    pushup(rt);
}
ll query(int L, int R, int l, int r, int rt){
    if(L <= l && R >= r){
        return sum[rt];
    }
    pushdown(rt, l, r);
    int m = (l + r) >> 1;
    ll ret = 0;
    if(L <= m)
        ret += query(L, R, l, m, rt << 1);
    if(R > m)
        ret += query(L, R, m + 1, r, rt << 1 | 1);
    return ret % p;
}
int main(){
    scanf("%lld%lld%lld", &n, &m, &p);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build(1, n, 1);
    while(m--){
        int op, x, y, k;
        scanf("%d%d%d", &op, &x, &y);
        if(op != 3) scanf("%d", &k);
        if(op == 1) update(x, y, 1, n, k, 1, 1);
        else if(op == 2) update(x, y, 1, n, k, 2, 1);
        else printf("%lld
", query(x, y, 1, n, 1));
    }
    return 0;
}