其实是个超傻逼的题目，但是交了20几发，就死在一个写惯了的小错误上

这种题目一看建两个标记就好了，

• \(tag1\)：表示区间加标记
• \(tag2\)：表示区间覆盖标记

那么下传方式很显然：

• 先下传 \(tag2\)，更新 \(tag2\) 和 \(v\) 和 \(tag1\)
• 然后下传 \(tag1\)，更新 \(v\) 和 \(tag1\)

就没了

然后我烦了一个很智障的错误:

我以为当 \(tag=0\) 时就是没有标记，

然后看一眼题目：\(|v|<=10^5\)，一脸懵逼去了

以后写线段树判断都要是 \(inf\) 才行

#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(#x".in", "r", stdin),freopen(#x".out", "w", stdout)
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 100010 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int MOD = 1000000007 ;
const double eps = 1e-7 ;
void print(int x) { cout << x << endl ; exit(0) ; }
void PRINT(string x) { cout << x << endl ; exit(0) ; }
void douout(double x){ printf("%lf\n", x + 0.0000000001) ; }int n, m, lst ;struct SegTree {int l, r, sz, v, tag1, tag2 ; // 一个求和标记，一个赋值标记#define ls(x) x << 1#define rs(x) x << 1 | 1#define l(x) tr[x].l#define r(x) tr[x].r#define sz(x) tr[x].sz#define v(x) tr[x].v#define tag1(x) tr[x].tag1#define tag2(x) tr[x].tag2
} tr[N << 2] ;void pushup(int x) {v(x) = v(ls(x)) + v(rs(x)) ;
}void pushdown(int x) {if (tag2(x) != iinf) {tag1(ls(x)) = tag1(rs(x)) = iinf ; // 清空？tag2(ls(x)) = tag2(rs(x)) = tag2(x) ;v(ls(x)) = sz(ls(x)) * tag2(x) ;v(rs(x)) = sz(rs(x)) * tag2(x) ;tag2(x) = iinf ;}if (tag1(x) != iinf) {tag1(ls(x)) = (tag1(ls(x)) == iinf ? 0 : tag1(ls(x))) + tag1(x) ;tag1(rs(x)) = (tag1(rs(x)) == iinf ? 0 : tag1(rs(x))) + tag1(x) ;v(ls(x)) += sz(ls(x)) * tag1(x) ;v(rs(x)) += sz(rs(x)) * tag1(x) ;tag1(x) = iinf ;}
}void build(int x, int l, int r) {l(x) = l, r(x) = r ;sz(x) = r(x) - l(x) + 1 ;tag1(x) = tag2(x) = iinf ;if (l == r) {v(x) = lst ;return ;}int mid = (l + r) >> 1 ;build(ls(x), l, mid) ;build(rs(x), mid + 1, r) ;pushup(x) ;
}void modify(int x, int l, int r, int c) {if (l <= l(x) && r(x) <= r) {tag1(x) = (tag1(x) == iinf ? 0 : tag1(x)) + c ;v(x) += sz(x) * c ;return ;}pushdown(x) ;int mid = (l(x) + r(x)) >> 1 ;if (l <= mid) modify(ls(x), l, r, c) ;if (mid < r) modify(rs(x), l, r, c) ;pushup(x) ;
}void cover(int x, int l, int r, int c) {if (l <= l(x) && r(x) <= r) {tag1(x) = iinf ; tag2(x) = c ;v(x) = c * sz(x) ;return ;}pushdown(x) ;int mid = (l(x) + r(x)) >> 1 ;if (l <= mid) cover(ls(x), l, r, c) ;if (mid < r) cover(rs(x), l, r, c) ;pushup(x) ;
}int query(int x, int l, int r) {if (l <= l(x) && r(x) <= r) return v(x) ;pushdown(x) ;int mid = (l(x) + r(x)) >> 1, ans = 0 ;if (l <= mid) ans += query(ls(x), l, r) ;if (mid < r) ans += query(rs(x), l, r) ;pushup(x) ;return ans ;
}signed main(){scanf("%lld%lld%lld", &n, &m, &lst) ;build(1, 1, n) ;while (m--) {int op, x, y, v ; scanf("%lld%lld%lld", &op, &x, &y) ;if (op == 0) {scanf("%lld", &v) ;modify(1, x, y, v) ;}else if (op == 1) {scanf("%lld", &v) ;cover(1, x, y, v) ;} else {printf("%lld\n", query(1, x, y)) ;}}return 0 ;
}/*
写代码时请注意：1.ll？数组大小，边界？数据范围？2.精度？3.特判？4.至少做一些
思考提醒：1.最大值最小->二分？2.可以贪心么？不行dp可以么3.可以优化么4.维护区间用什么数据结构？5.统计方案是用dp？模了么？6.逆向思维？
*/