开了 O2 也没用

#include <bits/stdc++.h>
#define int long long
#define lowbit(x) x&(-x)
using namespace std;
const int N = 1e6 + 5;
int n, m, len;
char op[N];
int nx[N], ny[N], b[N];
int c[N], c1[N], a[N];
inline void add(int x, int y, int t[])
{
	for (int i = x; i <= N; i += lowbit(i))
		t[i] += y;
}
inline int query(int x, int t[])
{
	int res = 0;
	for (int i = x; i >= 1; i -= lowbit(i))
		res += t[i];
	return res;
}
signed main()
{
	cin >> n >> m;
	for (int i = 1; i <= m; i++)
	{
	    scanf("%s%lld%lld", &op[i], &nx[i], &ny[i]);
		b[++len] = ny[i];
	}
	sort(b + 1, b + 1 + len);
	len = unique(b, b + len + 1) - b - 1;
	for (int i = 1; i <= m; i++)
	{
		if (ny[i]) ny[i] = lower_bound(b + 1, b + 1 + len, ny[i]) - b;//对于这行，如果不判断则是WA80
		int x = nx[i], y = ny[i];
		if (op[i] == 'U')
		{
			if (a[x] != 0)
			{
				add(a[x], -1, c);
				add(a[x], -b[a[x]], c1);
			}
			a[x] = y;
			add(a[x], 1, c);
			add(a[x], b[a[x]], c1);
		}
		else
		{
			int t = query(N, c) - query(y - 1, c);
			int t1 = (y != 0) ? query(y - 1, c1) : 0;
			if (t1 >= (x - t) * b[y]) cout << "TAK\n";
			else cout << "NIE\n";
		}
	}
	return 0;
}