XXII Open Cup, Grand Prix of Xi'an

Блог пользователя Qingyu

Автор Qingyu, история, 5 лет назад, По-английски

Could not find an official announcement, let's discuss problems here.

Обсуждение 2020-2021 Summer Petrozavodsk Camp, Day 6: XJTU Contest (XXII Open Cup, Grand Prix of XiAn)

+133

Qingyu
5 лет назад
11

Комментарии (11)

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moscow01

5 лет назад, скрыть # |

← Rev. 2 →

+14

How to solve Div.2 M(Colors and Pearls) and P(Number Sequence)?

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Qingyu

5 лет назад, скрыть # ^ |

+49

It looks like the problems for Div.2 were taken from some past ICPC Asia online contests...

M: A brute force solution is to denote $$$f_n$$$ as the answer for $$$a_{1} \cdots a_n$$$, then $$$f_n = \min_{0 \leq i \lt n} (f_i+v^2(i,n))$$$. Note that $$$f_n \leq n$$$, so we only need to keep the states that satisfy $$$v(i,n)\leq \sqrt n$$$. Time complexity is $$$\Theta(n \sqrt n)$$$.

Code

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 50;

int a[N], dp[N], f[N], g[N], lst[N];

int main()
{
	int n;
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	while (cin >> n)
	{
		for (int i = 1; i <= n; ++i) cin >> a[i];
		for (int i = 1; i <= n; ++i)
		{
			lst[i] = g[a[i]];
			g[a[i]] = i;
		}
		int tot = 0;
		memset(dp, 0x3f, n + 1 << 2);
		memset(f, 0, n + 1 << 2);
		memset(g, 0, n + 1 << 2);
		dp[0] = 0;
		for (int i = 1; i <= n; ++i)
		{
			int cnt = 0;
			g[++cnt] = i;
			for (int k = 1; k <= tot; ++k)
				if (a[f[k]] != a[i])
					g[++cnt] = f[k];
			for (int i = 1; i <= cnt; ++i) f[i] = g[i]; 
			tot = cnt;
			tot = std::min(tot, (int)sqrt(i) + 1);
			for (int k = 1; k <= tot; ++k)
				dp[i] = std::min(1ll * dp[i], dp[f[k + 1]] + 1ll * k * k);
		}
		printf("%d\n", dp[n]);
	}
}

N: For a pile of stones, we have $$$\mathcal G_n = \operatorname{mex}_{i+j \lt n} ( \mathcal G_i, \mathcal G_{i} \stackrel{*}{+} G_j )$$$, and you can proof $$$\mathcal G_n = *_n$$$ by induction. So the whole game is equivalent to the regular Nim game.

Code

#include <bits/stdc++.h>

using namespace std;

int main()
{
	int T, n;
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	while ((cin >> n))
	{
		int sum = 0, x;
		while (n--) { cin >> x; sum ^= x; }
		puts(sum ? "Win" : "Lose");
	}
}

O is just a simple breadth first search problem.

P: We can make each $$$a_i\oplus b_i$$$ become $$$(111\cdots 1)_2$$$ by greedily pairing. This is obviously the optimal answer.

Code

#include <bits/stdc++.h>

const int N = 1e5 + 50;
int a[N], b[N], n;
long long s, o, tot;

int main()
{
	while (~scanf("%d", &n))
	{
		s = 0;
		memset(b, -1, sizeof b);
		for (int i = 0; i <= n; ++i)
			scanf("%d", a + i);
		for (int i = n; i >= 0; --i)
			if (b[i] == -1)
			{
				tot = 1;
				while (tot <= i) tot <<= 1;
				tot--;
				o = tot ^ i;
				s += tot * 2;
				b[i] = o;
				b[o] = i;
			}
		printf("%lld\n", s);
		for (int i = 0; i <= n; ++i)
			printf(i == n ? "%d\n" : "%d ", b[a[i]]);
	}
	return 0;
}

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nor

5 лет назад, скрыть # |

+19

Thanks to the problemsetters, the problems were really nice (D in particular). How to solve C, G, J and K?

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Qingyu

5 лет назад, скрыть # ^ |

+65

There's an editorial prepared by the problemsetters.

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DrSwad

5 лет назад, скрыть # ^ |

← Rev. 2 →

+10

I had been looking for the editorial since the end of the contest, but couldn't seem to find it anywhere. Do you mind sharing how you found it? Most of the previous GP editorials were usually posted in the respective CF announcements, but this one doesn't seem to have any official announcement post.

Thanks for sharing the editorial btw.

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Qingyu

5 лет назад, скрыть # ^ |

+26

Well... In fact, the contest was used in Petrozavodsk Camp, and it was shared by the problemsetters during the camp.

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MiceMirzayanov

5 лет назад, скрыть # ^ |

-75

baaki sab ban gaya bhai?

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18o3

5 лет назад, скрыть # ^ |

Tere ban gaye kya?

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hos.lyric

5 лет назад, скрыть # |

+36

Thanks to problem F! I learned something new.

Here is a tricky case not in the judge tests (it's OK as I know it's not easy to create a strong cases):

1
10 9 100 3 -63 4

We have $$$E = \{ (3,1), (6,2), (9,3), (12,4), (15,5) \}$$$ and the answer should be 701423582.

My first attempt used quadtree-like DFS but I only checked the center of the ellipse and the boundary lattice points of the square to decide that they have any intersection. It is not correct for thin ellipses.

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