\(\text{Solution}\)

一看有區間賦值直接上 \(ODT\)

\(\text{Code}\)

#include <cstdio> 
#include <iostream>
#include <set>
#define re register
using namespace std;

int n;

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<node> T;
typedef set<node>::iterator IT;
inline IT split(int x)
{
	if (x > n) return T.end();
	IT 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)
{
	IT itr = split(r + 1), itl = split(l);
	T.erase(itl, itr), T.insert(node{l, r, v});
}
inline int Query(int l, int r, int v)
{
	IT itr = split(r + 1), itl = split(l); int res = 0;
	for(re IT it = itl; it != itr; ++it)
	if (it->v == v) res += it->r - it->l + 1;
	return res;
}

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

int main()
{
	read(n);
	for(re int i = 1, x; i <= n; i++) read(x), T.insert((node){i, i, x});
	for(re int i = 1, l, r, v; i <= n; i++)
		read(l), read(r), read(v), printf("%d\n", Query(l, r, v)), assign(l, r, v);
}