A

取最大值, 取的值的数量是固定的

对长度n为偶数的数组, 一定是 n / 2

对奇数, 如果取值在奇数位, 那么是(n + 1) / 2, 取值在偶数位则是n / 2, 因此能取奇数位取奇数位

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int N = 200010;

int n, m, k, q, ans;

int a[N], f[N][2];

void solve()
{
	cin >> n;
	int maxx = 0;
	int ans = n / 2;
	for (int i = 1; i <= n; i ++ )
	{
		cin >> a[i];
		maxx = max(a[i], maxx);
		
	}
	
	for	(int i = 1; i <= n; i ++ )
	{
		if(a[i] == maxx && i % 2 == 1)
		{
			ans = (n + 1) / 2;
		}
	}
	
	
	cout << maxx + ans << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	cin >> T;
	while (T -- )
	{
		solve();
	}
}

B

现在我们设第i 个点左面有b个线段端点, 右面有c个线段端点

如果这个点不是端点, 那么有b * c 个点经过它

如果是的话, 有b * c + b + c个点经过它(考虑它本身与其它端点相交)

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int N = 200010;

int n, m, k, q, ans;

int a[N], b[N], c[N];

void solve()
{
	cin >> n >> q;
	
	for (int i = 1; i <= n; i ++ )
	{
		cin >> a[i];
	}
	
	
	map<int, int> mp;
	
	for (int i = 1; i < n; i ++ )
	{
		mp[i * (n - i)] += a[i + 1] - a[i] - 1;
//		cout << "i : " << i << " " << i * (n - i) << " " << mp[i * (n - i)] << "\n";
	}
	
	for	(int i = 1; i <= n; i ++ )
	{
		mp[n - i + (i - 1) * (n - i + 1)] += 1;
	}
	
	for (int i = 1; i <= q; i ++ )
	{
		int k;
		cin >> k;
		cout << mp[k] << " ";
	}
	cout << "\n";
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	cin >> T;
	while (T -- )
	{
		solve();
	}
}

C

这道题比较思维

maxx 取同一种牌出现的最多的数目, sum 取总牌数

检查 i 可不可以作为一组deck的长度即可. 而检查只需要判断sum + i 是否大于maxx * i 和 (sum + i - 1) / i * i 这两者即可

这是因为如果分组, 那么至少分maxx组, 也至少分(sum + i - 1) / i 组

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 200010;

int n, m;
int a[N];

void solve()
{
	cin >> n >> m;
	int sum = 0, maxx = 0;
	for(int i = 1; i <= n; i ++ )
	{
		cin >> a[i];
		maxx = max(maxx, a[i]);
		sum += a[i];
	}
	
	int ans = 1;
	
	for(int i = 2; i <= n; i ++ )
	{
		int minn = max(maxx * i, (sum + i - 1) / i * i);
		if(minn <= sum + m)
		{
			ans = max(i, ans);
		}
	}
	cout << ans << "\n";
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	cin >> T;
	while (T -- )
	{
		solve();
	}
}