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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#369745#2827. AutobiographyPetroTarnavskyi#WA 1ms3552kbC++202.1kb2024-03-28 17:21:422024-03-28 17:21:43

Judging History

你现在查看的是最新测评结果

  • [2024-03-28 17:21:43]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:3552kb
  • [2024-03-28 17:21:42]
  • 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;

struct SegTree
{
	VI lazy, cnt;
	vector<LL> sum, all;
	void init(int _n)
	{
		int n = 1;
		while(n <= _n) n *= 2;
		n *= 2;
		lazy.resize(n);
		cnt.resize(n);
		sum.resize(n);
		all.resize(n);
	}
	void combine(int v)
	{
		cnt[v] = cnt[2 * v + 1] + cnt[2 * v + 2];
		sum[v] = sum[2 * v + 1] + sum[2 * v + 2];
		all[v] = all[2 * v + 1] + all[2 * v + 2];
	}
	void build(int v, int tl, int tr, const string& s)
	{
		if(tl + 1 == tr)
		{
			cnt[v] = 0;
			all[v] = tl + 1;
			if(s[tl] == '1')
			{
				cnt[v] = 1;
				sum[v] = all[v];
			}
			return;
		}
		int tm = (tl + tr) / 2;
		build(2 * v + 1, tl, tm, s);
		build(2 * v + 2, tm, tr, s);
		
		combine(v);
	}
	void push(int v, int tl, int tr)
	{
		if(!lazy[v])
			return;
		lazy[v] = 0;
		sum[v] = all[v] - sum[v];
		cnt[v] = (tr - tl) - cnt[v];
		
		if(tl + 1 == tr)
			return;
		lazy[2 * v + 1] ^= 1;	
		lazy[2 * v + 2] ^= 1;	
	}
	void upd(int v, int tl, int tr, int l, int r)
	{
		push(v, tl, tr);
		if(tr <= l || r <= tl)
			return;
		if(l <= tl && tr <= r)
		{
			lazy[v] = 1;
			push(v, tl, tr);
			return;
		}
		int tm = (tl + tr) / 2;
		upd(2 * v + 1, tl, tm, l, r);
		upd(2 * v + 2, tm, tr, l, r);
		
		combine(v);
	}
};



int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int n, q;
	while(cin >> n >> q)
	{
		string s;
		cin >> s;
		SegTree T;
		T.init(n);
		T.build(0, 0, n, s);
		while(q--)
		{
			int l, r;
			cin >> l >> r;
			T.upd(0, 0, n, l - 1, r);
			assert(T.lazy[0] == 0);
			
			LL sum = T.sum[0];
			LL cnt = T.cnt[0];
		
			//cout << sum << " " << cnt << "\n";
			cout << 2 * sum - cnt * cnt << "\n";
		}
	}
	
	
	return 0;
}

详细

Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 3552kb

input:

5 4
bbobo
1 3
2 3
3 4
4 5
4 6
bobo
1 2
1 3
1 4
2 3
2 4
3 4
4 0
bobo

output:

3
1
7
9
2
5
5
7
2
4

result:

wrong answer 1st lines differ - expected: '2', found: '3'