Codeforces - 1166E - The LCMs Must be Large

地址

http://codeforces.com/contest/1166/problem/E

原文地址

https://www.lucien.ink/archives/434

题目

Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?

There are nnn stores numbered from 111 to nnn in Nlogonia. The iii-th of these stores offers a positive integer aia_iai.

Each day among the last mmm days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.

Dora considers Swiper to be her rival, and she considers that she beat Swiper on day iii if and only if the least common multiple of the numbers she bought on day iii is strictly greater than the least common multiple of the numbers that Swiper bought on day iii.

The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.

However, Dora forgot the values of aia_iai. Help Dora find out if there are positive integer values of aia_iai such that she beat Swiper on every day. You don’t need to find what are the possible values of aia_iai though.

Note that it is possible for some values of aia_iai to coincide in a solution.

题意

有 nnn 种石头，每个石头有一个未知的权值，有 mmm 天，每天一个整数 kkk 代表这一天拿了 kkk 个石头，接下来是 kkk 个整数，代表石头的下标。

问是否存在一种分配权值的方案，使得每一天拿的所有石头权值的 LCMLCMLCM 都严格大于补集中所有石头权值的 LCMLCMLCM 。

题解

反证法，考虑某一天拿的集合为 DiD_iDi ，补集为 SiS_iSi ，另一天的集合为 DjD_jDj ，补集为 SjS_jSj ，且 SiS_iSi 严格等于 DjD_jDj 。假设题设成立，那么有：

LCM(Di)>LCM(Si)=LCM(Dj)>LCM(Sj)=LCM(Di)LCM(D_i) > LCM(S_i) = LCM(D_j) > LCM(S_j) = LCM(D_i)LCM(Di)>LCM(Si)=LCM(Dj)>LCM(Sj)=LCM(Di)

显然矛盾，所以不成立。

考虑另一个问题，假设 AAA 是 BBB 的子集，那么一定存在 LCM(B)≥LCM(A)LCM(B) \ge LCM(A)LCM(B)≥LCM(A) 而且 LCM(B)mod  LCM(A)=0LCM(B) \mod LCM(A) = 0LCM(B)modLCM(A)=0 ，假设 DiD_iDi 和 DjD_jDj 的交集为空，那么 DjD_jDj 一定为 SiS_iSi 的真子集，同理，DiD_iDi 也是 SjS_jSj 的真子集，那么：

LCM(Di)>LCM(Si)≥LCM(Dj)>LCM(Sj)≥LCM(Di)LCM(D_i) > LCM(S_i) \ge LCM(D_j) > LCM(S_j) \ge LCM(D_i)LCM(Di)>LCM(Si)≥LCM(Dj)>LCM(Sj)≥LCM(Di)

显然也矛盾，即：任意两天所拿的石头都必须有交集 (1)。

2019年5月22日补充

虽然不满足条件 (1) 一定不能赢，但满足条件 (1) 就一定可以赢吗？答案是不一定。但我们可以通过合理的构造来使得当满足条件 (1) 时是一定全胜的。

如何构造出来一组解呢？根据结论 (1) ，一个显然的想法就是将权值全都集中在有交集的那部分石头上，剩下的全为 111 ，这样一来我方取到的石头的 LCM 一定是严格大于对面的。

更详细一些，我们取出 mmm 个互不相同的素数 p1,p2,…,pm−1,pm{p_1, p_2, \dots, p_{m - 1}, p_m}p1,p2,…,pm−1,pm ，令 nnn 个石头的权值为 a1,a2,a3,…,an−1,an{a_1, a_2, a_3, \dots, a_{n - 1}, a_n}a1,a2,a3,…,an−1,an ，初始时 aia_iai 均为 111 。假设我们在第 jjj 天访问了 aia_iai ，那么我们就令 ai=ai⋅pja_i = a_i \cdot p_jai=ai⋅pj ，这样一来 pjp_jpj 就是 aia_iai 的一个因子。

下面来证明这样分配一定是必胜的。我们知道在第 iii 天，我们访问了一个在第 jjj 天访问过的石头，而且这个石头的权值一定是 pjp_jpj 的倍数。所以对于每一天 iii ，LCM(Di)=p1p2…pmLCM(D_i) = p_1p_2\dots p_mLCM(Di)=p1p2…pm 。另一方面，SiS_iSi 并不包含 pip_ipi 这个因子，因为 pip_ipi 都在 DiD_iDi 里，所以 LCM(Si)LCM(S_i)LCM(Si) 严格小于 LCM(Di)LCM(D_i)LCM(Di) 。

mmm 的范围只有 505050 ，所以直接 O(nm2)O(nm^2)O(nm2) 暴力判断一下即可。

代码

https://pasteme.cn/8174

#include <bits/stdc++.h>
const int maxm = 57, maxn = int(1e4) + 7;
int m, n;
bool vis[maxm][maxn];
int main() {scanf("%d%d", &m, &n);for (int i = 1, cnt, buf; i <= m; i++) {scanf("%d", &cnt);while (cnt--) scanf("%d", &buf), vis[i][buf] = true;}for (int i = 1; i <= m; i++) {for (int j = 1; j <= m; j++) {bool flag = false;for (int k = 1; k <= n && !flag; k++) {if (vis[i][k] && vis[j][k]) flag = true;}if (!flag) return 0 * puts("impossible");}}puts("possible");return 0;
}