2025.1.19——1300
2025.1.19——1300
A 1300
You are given an array \(a\) consisting of \(n\) positive integers. You can perform the following operation on it:
- Choose a pair of elements \(a_i\) and \(a_j\) (\(1 \le i, j \le n\) and \(i \neq j\));
- Choose one of the divisors of the integer \(a_i\), i.e., an integer \(x\) such that \(a_i \bmod x = 0\);
- Replace \(a_i\) with \(\frac{a_i}{x}\) and \(a_j\) with \(a_j \cdot x\).
Determine whether it is possible to make all elements in the array the same by applying the operation a certain number of times (possibly zero).
For example, let's consider the array \(a\) = [\(100, 2, 50, 10, 1\)] with \(5\) elements. Perform two operations on it:
- Choose \(a_3 = 50\) and \(a_2 = 2\), \(x = 5\). Replace \(a_3\) with \(\frac{a_3}{x} = \frac{50}{5} = 10\), and \(a_2\) with \(a_2 \cdot x = 2 \cdot 5 = 10\). The resulting array is \(a\) = [\(100, 10, 10, 10, 1\)];
- Choose \(a_1 = 100\) and \(a_5 = 1\), \(x = 10\). Replace \(a_1\) with \(\frac{a_1}{x} = \frac{100}{10} = 10\), and \(a_5\) with \(a_5 \cdot x = 1 \cdot 10 = 10\). The resulting array is \(a\) = [\(10, 10, 10, 10, 10\)].
After performing these operations, all elements in the array \(a\) become equal to \(10\).
Input
The first line of the input contains a single integer \(t\) (\(1 \le t \le 2000\)) — the number of test cases.
Then follows the description of each test case.
The first line of each test case contains a single integer \(n\) (\(1 \le n \le 10^4\)) — the number of elements in the array \(a\).
The second line of each test case contains exactly \(n\) integers \(a_i\) (\(1 \le a_i \le 10^6\)) — the elements of the array \(a\).
It is guaranteed that the sum of \(n\) over all test cases does not exceed \(10^4\).
B 1300
You have \(n\) sets of integers \(S_{1}, S_{2}, \ldots, S_{n}\). We call a set \(S\) attainable, if it is possible to choose some (possibly, none) of the sets \(S_{1}, S_{2}, \ldots, S_{n}\) so that \(S\) is equal to their union\(^{\dagger}\). If you choose none of \(S_{1}, S_{2}, \ldots, S_{n}\), their union is an empty set.
Find the maximum number of elements in an attainable \(S\) such that \(S \neq S_{1} \cup S_{2} \cup \ldots \cup S_{n}\).
\(^{\dagger}\) The union of sets \(A_1, A_2, \ldots, A_k\) is defined as the set of elements present in at least one of these sets. It is denoted by \(A_1 \cup A_2 \cup \ldots \cup A_k\). For example, \(\{2, 4, 6\} \cup \{2, 3\} \cup \{3, 6, 7\} = \{2, 3, 4, 6, 7\}\).
Input
Each test contains multiple test cases. The first line contains the number of test cases \(t\) (\(1 \le t \le 100\)). The description of the test cases follows.
The first line of each test case contains a single integer \(n\) (\(1 \le n \le 50\)).
The following \(n\) lines describe the sets \(S_1, S_2, \ldots, S_n\). The \(i\)-th of these lines contains an integer \(k_{i}\) (\(1 \le k_{i} \le 50\)) — the number of elements in \(S_{i}\), followed by \(k_{i}\) integers \(s_{i, 1}, s_{i, 2}, \ldots, s_{i, k_{i}}\) (\(1 \le s_{i, 1} < s_{i, 2} < \ldots < s_{i, k_{i}} \le 50\)) — the elements of \(S_{i}\).
------------------------思考------------------------
-
数学+思维/入手点
A
- 本质操作是移动因子/质因子。
- 从结果来看，分解质因数后发现每个质因子出现的次数都必须为 \(n\) 的倍数。具有充分必要性。
B
- 入手点是枚举每个数：当此数不在最后集合中时，暴力统计有多少元素没有受到影响。最后坏情况大约 6e7 , 600ms 。
- 搞了半天数据结构维护，原来暴力就行。
------------------------代码------------------------
A
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // attention: interactive/debug
#define el cout << endl
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC, ST) \
for (int i = ST; i < VEC.size(); i++) \
cout << VEC[i] << ' '; \
el;
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(10);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int n;
cin >> n;
map<int, int> cnt;
for (int i = 0; i < n; i++)
{
int x;
cin >> x;
for (int j = 2; j <= x / j && x; j++)
while (x % j == 0)
{
cnt[j]++;
x /= j;
}
if (x - 1)
cnt[x]++;
}
// for (auto [x, has] : cnt)
// bug2(x, has);
bool f = 1;
for (auto [x, has] : cnt)
if (has % n)
f = 0;
cout << (f ? "YES" : "NO");
el;
}
B
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // attention: interactive/debug
#define el cout << endl
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC, ST) \
for (auto i = VEC.begin() + ST; i != VEC.end(); i++) \
cout << *i << ' '; \
el;
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(10);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int n;
cin >> n;
vector<set<int>> has(n);
set<int> t;
for (auto &s : has)
{
int cnt;
cin >> cnt;
for (int i = 0; i < cnt; i++)
{
int x;
cin >> x;
s.insert(x);
t.insert(x);
}
}
n = t.size();
int res = 0;
for (auto x : t)
{
set<int> left;
for (auto s : has)
{
auto it = s.lower_bound(x);
// bug(*it);
if (it == s.end() || *it != x)
for (auto k : s)
left.insert(k);
}
// bug2(x, left.size());
res = max(res, (int)left.size());
}
cout << res << endl;
}
// void _()
// {
// int n;
// cin >> n;
// map<int, map<int, int>> has;
// for (int i = 0; i < n; i++)
// {
// int cnt;
// cin >> cnt;
// set<int> s;
// for (int i = 0; i < cnt; i++)
// {
// int x;
// cin >> x;
// s.insert(x);
// }
// for (auto x : s)
// for (auto y : s)
// has[x][y]++;
// }
// map<int, int> w;
// for (auto [x, hs] : has)
// {
// int c = 0;
// for (auto [y, cnt] : hs)
// c += cnt == 1;
// w[x] = c;
// }
// int res = 0;
// for (auto [x, v] : w)
// res = max(res, (int)w.size() - v), bug2(x, v);
// cout << res << endl;
// }
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