2024.12.21 周六
Q1. 1000
Lottery "Three Sevens" was held for m m m days. On day i i i, n i n_i ni people with the numbers a i , 1 , ... , a i , n i a_{i, 1}, \ldots, a_{i, n_i} ai,1,...,ai,ni participated in the lottery.
It is known that in each of the m m m days, only one winner was selected from the lottery participants. The lottery winner on day i i i was not allowed to participate in the lottery in the days from i + 1 i+1 i+1 to m m m.
Unfortunately, the information about the lottery winners has been lost. You need to find any possible list of lottery winners on days from 1 1 1 to m m m or determine that no solution exists.
Q2. 1000
Kristina has a string s s s of length n n n, consisting only of lowercase and uppercase Latin letters. For each pair of lowercase letter and its matching uppercase letter, Kristina can get 1 1 1 burl. However, pairs of characters cannot overlap, so each character can only be in one pair.
For example, if she has the string s s s = "aAaaBACacbE", she can get a burl for the following character pairs:
- s 1 s_1 s1 = "a" and s 2 s_2 s2 = "A"
- s 4 s_4 s4 = "a" and s 6 s_6 s6 = "A"
- s 5 s_5 s5 = "B" and s 10 s_{10} s10 = "b"
- s 7 s_7 s7= "C" and s 9 s_9 s9 = "c"
Kristina wants to get more burles for her string, so she is going to perform no more than k k k operations on it. In one operation, she can:
- either select the lowercase character s i s_i si ( 1 ≤ i ≤ n 1 \le i \le n 1≤i≤n) and make it uppercase.
- or select uppercase character s i s_i si ( 1 ≤ i ≤ n 1 \le i \le n 1≤i≤n) and make it lowercase.
For example, when k k k = 2 and s s s = "aAaaBACacbE" it can perform one operation: choose s 3 s_3 s3 = "a" and make it uppercase. Then she will get another pair of s 3 s_3 s3 = "A" and s 8 s_8 s8 = "a"
Find maximum number of burles Kristina can get for her string.
Q3. 1000
You are given a two-dimensional plane, and you need to place n n n chips on it.
You can place a chip only at a point with integer coordinates. The cost of placing a chip at the point ( x , y ) (x, y) (x,y) is equal to ∣ x ∣ + ∣ y ∣ |x| + |y| ∣x∣+∣y∣ (where ∣ a ∣ |a| ∣a∣ is the absolute value of a a a).
The cost of placing n n n chips is equal to the maximum among the costs of each chip.
You need to place n n n chips on the plane in such a way that the Euclidean distance between each pair of chips is strictly greater than 1 1 1, and the cost is the minimum possible.
------------------------独自思考分割线------------------------
Q1 Q3有些意思。1000分的题先到这，位运算，构造，数论，三者最难。
A1.
- 发现制约关系：当前答案的数后面的数组都不能出现。从后向前扫，未标记的数就可以作为答案。
- 这是必要性，需要证明一下充分性。
A2.
- 本质就是大小写匹配的次数，剩下的自己匹配自己(需要代价)。
- 前一天扫描贪心匹配wa了，现在复习也没时间去找bug。先这样了
A3.
- 被卡住了 还有些疑问，去学vue写报告了，先这样
------------------------代码分割线------------------------
A1.
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // 交互/调试 关
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
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(6);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int m;
cin >> m;
vector<vector<int>> a(m);
vector<int> vis(50010);
for (auto &v : a)
{
int n;
cin >> n;
while (n--)
{
int x;
cin >> x;
v.push_back(x);
}
}
vector<int> res;
for (int i = m - 1; i >= 0; i--)
{
auto v = a[i];
int f = 0;
for (auto t : v)
{
if (!vis[t] && !f)
{
f = 1;
res.push_back(t);
}
vis[t] = 1;
}
}
if (res.size() - m)
res.assign(1, -1);
reverse(res.begin(), res.end());
for (auto v : res)
cout << v << " ";
cout << endl;
}
A2.
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // 交互/调试 关
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
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(6);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int m;
cin >> m;
vector<vector<int>> a(m);
vector<int> vis(50010);
for (auto &v : a)
{
int n;
cin >> n;
while (n--)
{
int x;
cin >> x;
v.push_back(x);
}
}
vector<int> res;
for (int i = m - 1; i >= 0; i--)
{
auto v = a[i];
int f = 0;
for (auto t : v)
{
if (!vis[t] && !f)
{
f = 1;
res.push_back(t);
}
vis[t] = 1;
}
}
if (res.size() - m)
res.assign(1, -1);
reverse(res.begin(), res.end());
for (auto v : res)
cout << v << " ";
cout << endl;
}
A3.
#include <bits/stdc++.h>
#define int long long //
#define endl '\n' // 交互/调试 关
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
void _();
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(6);
int T = 1;
cin >> T;
while (T--)
_();
return 0;
}
void _()
{
int n;
cin >> n;
int res = ceil(sqrtl(n)) - 1;
cout << res << endl;
}
// void _()
// {
// int n;
// cin >> n;
// int qrt_n = sqrt(n);
// int k = qrt_n / 2 + (qrt_n * qrt_n != n);
// int res = k - 1 << 1 | 1;
// if ((k - 1) * (k - 1) << 2 == n - 1)
// res--;
// if (n == 1)
// res = 0;
// // bug(k);
// cout << res << endl;
// }