2024.12.12 周四
2024.12.12 周四
Q1. 1000
You have an array \(a\) of \(n\) integers.
You can no more than once apply the following operation: select three integers \(i\), \(j\), \(x\) (\(1 \le i \le j \le n\)) and assign all elements of the array with indexes from \(i\) to \(j\) the value \(x\). The price of this operation depends on the selected indices and is equal to \((j - i + 1)\) burles.
What is the least amount of burles you need to spend to make all the elements of the array equal?
Q2. 1000
You are given a positive integer \(n\).
Find a permutation\(^\dagger\) \(p\) of length \(n\) such that there do not exist two distinct indices \(i\) and \(j\) such that \(p_i\) divides \(p_j\) and \(p_{i+1}\) divides \(p_{j+1}\).
Under the constraints of this problem, it can be proven that at least one \(p\) exists.
Q3. 1000
Given an array \(a\) of \(n\) integers, an array \(b\) of \(m\) integers, and an even number \(k\).
Your task is to determine whether it is possible to choose exactly \(\frac{k}{2}\) elements from both arrays in such a way that among the chosen elements, every integer from \(1\) to \(k\) is included.
Q4. 1000
A certain number \(1 \le x \le 10^9\) is chosen. You are given two integers \(a\) and \(b\), which are the two largest divisors of the number \(x\). At the same time, the condition \(1<=a<b<x\) is satisfied.
For the given numbers \(a\), \(b\), you need to find the value of \(x\).
Q5. 1000
You are given a binary string \(s\).
You can perform two types of operations on \(s\):
- delete one character from \(s\). This operation costs \(1\) coin;
- swap any pair of characters in \(s\). This operation is free (costs \(0\) coins).
You can perform these operations any number of times and in any order.
Let's name a string you've got after performing operations above as \(t\). The string \(t\) is good if for each \(i\) from \(1\) to \(|t|\) \(t_i \neq s_i\) (\(|t|\) is the length of the string \(t\)). The empty string is always good. Note that you are comparing the resulting string \(t\) with the initial string \(s\).
What is the minimum total cost to make the string \(t\) good?
------------------------独自思考分割线------------------------
-
Q2构造 Q4数论 比较有意思
A1.
- 发现只有3种情况，找边界枚举即可。
A2.
- 倍数就一定相差至少一倍。
- 如何构造使相邻的2个数都无法找到倍数，1~n存在一半的数无法找到其倍数，因此考虑存在倍数与不存在倍数的数相邻。如：1 n 2 n-1 3 n-2...
A3.
- 变量维护两数组选的数的个数，map维护两数组有的数。
- 遍历1~k：考虑独有的，直接加入，若不能加入/都没有则无解。共有的不需要考虑，因为只要二者选的数次数满足，最后一定可以满足。
A4.
- 对 \(x\) 因式分解，b=x/p1,a=x/p1 \(or\) x/p2，...嗯..是道好题。
A5.
- 任意交换代表任意安排，贪心从前向后构造，直到不能构造，删除剩下的。
------------------------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 n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++)
cin >> a[i];
int res = n;
int l = 1;
for (; l + 1 <= n && a[l + 1] == a[1]; l++)
;
res = min(res, n - l);
int r = n;
for (; r > 1 && a[r - 1] == a[n]; r--)
;
res = min(res, r - 1);
if (a[1] == a[n])
{
if (l >= r)
res = 0;
else
res = min(res, r - l - 1);
}
cout << res << 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 n;
cin >> n;
int l = 1, r = n;
int f = 1;
while (l <= r)
{
if (f)
cout << l << ' ', l++;
else
cout << r << ' ', r--;
f ^= 1;
}
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, m, k;
cin >> n >> m >> k;
map<int, int> ka, kb;
for (int i = 0; i < n; i++)
{
int x;
cin >> x;
ka[x] = 1;
}
for (int i = 0; i < m; i++)
{
int x;
cin >> x;
kb[x] = 1;
}
bool ans = 1;
int cnta = 0, cntb = 0;
for (int i = 1; i <= k; i++)
if (ka[i] + kb[i] == 1)
{
int f = 0;
if (ka[i] && cnta < k / 2)
cnta++, f = 1;
if (kb[i] && cntb < k / 2)
f = 1, cntb++;
if (!f)
ans = f;
}
else if (ka[i] + kb[i] == 0)
ans = 0;
cout << (ans ? "YES" : "NO") << endl;
}
// void _()
// {
// int n, m, k;
// cin >> n >> m >> k;
// map<int, int> ka, kb;
// for (int i = 0; i < n; i++)
// {
// int x;
// cin >> x;
// ka[x]++;
// }
// for (int i = 0; i < m; i++)
// {
// int x;
// cin >> x;
// kb[x]++;
// }
// set<int> K;
// int cnta = 0, cntb = 0;
// for (int i = 1; i <= k; i++)
// if (ka[i] + kb[i] == 1)
// {
// if (ka[i] && cnta < k / 2)
// K.insert(i), cnta++;
// if (kb[i] && cntb < k / 2)
// K.insert(i), cntb++;
// }
// for (int i = 1; i <= k; i++)
// if (ka[i] + kb[i] > 1)
// {
// if (cnta < k / 2)
// {
// if (ka[i])
// {
// K.insert(i), cnta++;
// continue;
// }
// }
// if (cntb < k / 2)
// {
// if (kb[i])
// {
// K.insert(i), cntb++;
// continue;
// }
// }
// }
// cout << (K.size() == k ? "YES" : "NO") << endl;
// }
A4.
#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 a, b;
cin >> a >> b;
auto get = [](int n)
{
for (int i = 2; i <= n / i; i++)
if (n % i == 0)
return i;
return n;
};
int res = b * b / a;
if (b % a)
res = b * get(a);
cout << res << endl;
}
A5
#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 _()
{
string s;
cin >> s;
int n = s.size();
s = " " + s;
int cnt0 = 0;
for (int i = 1; i <= n; i++)
cnt0 += s[i] == '0';
int cnt1 = n - cnt0;
int res = 0;
// bug2(cnt0, cnt1);
for (int i = 1; i <= n; i++)
{
if (s[i] == '0')
{
if (!cnt1)
{
res = n - (i - 1);
break;
}
cnt1--;
}
else
{
if (!cnt0)
{
res = n - (i - 1);
break;
}
cnt0--;
}
}
cout << res << endl;
}
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