\(\text{Naive Solition}\)

当然是 \(ODT\) 暴力啦
\(Luogu\) 煞费苦心加强了数据，于是就过不了了。。。
不过 \(LibreOJ\) 上可以过

#include  
#include 
#include 
#define re register
#define iter set::iterator
using namespace std;

inline void read(int &x)
{
	x = 0; char ch = getchar();
	for(; !isdigit(ch); ch = getchar());
	for(; isdigit(ch); x = (x<<3) + (x<<1) + (ch^48), ch = getchar());
}

int n, m;
struct node{
	int l, r; mutable int v;
	inline node(int l, int r, int v):l(l), r(r), v(v){};
	inline bool operator < (const node &a) const {return l < a.l;}
};
set T;

inline iter split(int x)
{
	if (x > n) return T.end();
	iter it = --T.upper_bound(node{x, 0, 0});
	if (it->l == x) return it;
	int l = it->l, r = it->r, v = it->v;
	T.erase(it), T.insert(node{l, x - 1, v});
	return T.insert(node{x, r, v}).first;
}
inline void assign(int l, int r, int v)
{
	iter itr = split(r + 1), itl = split(l);
	T.erase(itl, itr), T.insert(node{l, r, v});
}
inline int QuerySum(int l, int r)
{
	iter itr = split(r + 1), itl = split(l); int sum = 0;
	for(re iter it = itl; it != itr; ++it) sum += it->v * (it->r - it->l + 1);
	return sum;
}
inline void Modify(int l, int r, int x, int y)
{
	int sum = QuerySum(l, r); assign(l, r, 0);
	iter itr = split(y + 1), itl = split(x);
	for(re iter it = itl; it != itr && sum; ++it)
	if (it->v == 0)
	{
		int len = it->r - it->l + 1;
		if (len <= sum) sum -= len, it->v = 1;
		else assign(it->l, it->l + sum - 1, 1), sum = 0;
	}
}
inline int QueryLong(int l, int r)
{
	iter itr = split(r + 1), itl = split(l); int mx = 0, sum = 0;
	for(re iter it = itl; it != itr; ++it)
	{
		if (it->v == 0) sum += (it->r - it->l + 1);
		else mx = (mx < sum ? sum : mx), sum = 0;
	}
	return (mx < sum ? sum : mx);
}

int main()
{
	read(n), read(m), T.insert(node{1, n, 1});
	for(int op, l, r, x, y; m; --m)
	{
		read(op);
		if (op == 0) read(l), read(r), assign(l, r, 0);
		else if (op == 1) read(l), read(r), read(x), read(y), Modify(l, r, x, y);
		else read(l), read(r), printf("%d\n", QueryLong(l, r));
	}
}