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#208621#6677. Puzzle: SashiganePetroTarnavskyi#WA 0ms5548kbC++173.1kb2023-10-09 19:26:082023-10-09 19:26:08

Judging History

你现在查看的是最新测评结果

  • [2023-10-09 19:26:08]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:5548kb
  • [2023-10-09 19:26:08]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;

const int mod = 998244353;
const int N = 1 << 18;

void updAdd(int& x, int val)
{
	x += val;
	if (x >= mod)
		x -= mod;
}

int add(int a, int b)
{
	return a + b < mod ? a + b : a + b - mod;
}

int mult(int a, int b)
{
	return (LL)a * b % mod;
}

struct SegTree
{
	int t[N];
	int lazy[N];
	void build(int v, int tl, int tr)
	{
		t[v] = 0;
		lazy[v] = 1;
		if (tr == tl + 1)
			return;
		int tm = (tl + tr) / 2;
		build(2 * v + 1, tl, tm);
		build(2 * v + 2, tm, tr);
	}
	void push(int v)
	{
		if (2 * v < N)
		{
			FOR(i, 1, 3)
				lazy[2 * v + i] = mult(lazy[2 * v + i], lazy[v]);
		}
		t[v] = mult(t[v], lazy[v]);
		lazy[v] = 1;
	}
	void updAssign(int v, int tl, int tr, int pos, int val)
	{
		push(v);
		if (tr == tl + 1)
		{
			t[v] = val;
			return;
		}
		int tm = (tl + tr) / 2;
		if (pos < tm)
			updAssign(2 * v + 1, tl, tm, pos, val);
		else
			updAssign(2 * v + 2, tm, tr, pos, val);
		t[v] = add(t[2 * v + 1], t[2 * v + 2]);
	}
	void updMult(int v, int tl, int tr, int l, int r)
	{
		push(v);
		if (tr <= l || r <= tl)
			return;
		if (l <= tl && tr <= r)
		{
			lazy[v] = 2;
			push(v);
			return;
		}
		int tm = (tl + tr) / 2;
		updMult(2 * v + 1, tl, tm, l, r);
		updMult(2 * v + 2, tm, tr, l, r);
		t[v] = add(t[2 * v + 1], t[2 * v + 2]);
	}
	int query(int v, int tl, int tr, int l, int r)
	{
		push(v);
		if (tr <= l || r <= tl)
			return 0;
		if (l <= tr && tr <= r)
			return t[v];
		int tm = (tl + tr) / 2;
		return add(query(2 * v + 1, tl, tm, l, r), query(2 * v + 2, tm, tr, l, r));
	}
} st[2];

void solve()
{
	int n;
	cin >> n;
	vector<PII> segs[2];
	FOR(i, 0, 2)
		segs[i] = {{0, 0}};
	FOR(i, 0, n)
	{
		int l, r, c;
		cin >> l >> r >> c;
		segs[c].PB({r, l});
	}
	vector<int> dp[2];
	FOR(i, 0, 2)
	{
		sort(ALL(segs[i]));
		dp[i].resize(SZ(segs[i]));
		dp[i][0] = 1;
		st[i].build(0, 0, SZ(dp[i]));
		st[i].updAssign(0, 0, SZ(dp[i]), 0, 1);
	}
	int ptr[2] = {1, 1};
	int ans = 1;
	while (ptr[0] < SZ(dp[0]) || ptr[1] < SZ(dp[1]))
	{
		int k = ptr[1] < SZ(dp[1]) && (ptr[0] == SZ(dp[0]) || segs[1][ptr[1]] < segs[0][ptr[0]]);
		int j = lower_bound(ALL(segs[k ^ 1]), MP(segs[k][ptr[k]].S, 0)) - segs[k ^ 1].begin();
		int cur = st[k ^ 1].query(0, 0, SZ(dp[k ^ 1]), 0, j);
		st[k].updAssign(0, 0, SZ(dp[k]), ptr[k], cur);
		st[k ^ 1].updAssign(0, 0, SZ(dp[k ^ 1]), 0, j);
		ptr[k]++;
		updAdd(ans, cur);
		cerr << k << " " << ptr[k] - 1 << " " << segs[k][ptr[k] - 1].S << " " << segs[k][ptr[k] - 1].F << " " << cur << endl;
	}
	cout << ans << "\n";
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(15);
	int t;
	cin >> t;
	while(t--)
		solve();
	return 0;	
}

詳細信息

Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 5548kb

input:

5 3 4

output:

4
4
4
4
4

result:

wrong answer YES or NO expected in answer, but 4 found. (test case 1)