CF EDU

A.

易得。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <cstdlib>
#include <string>
using namespace std;

#define ll long long

const int N = 2e5 + 5, INF = 2e9, MOD = 1e9 + 7;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, t;

int main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	cin >> t;
	while (t--)
	{
		string s;
		cin >> s;
		if (s.size() == 2 || s.size() == 3)
		{
			printf("No\n");
			continue;
		}
		map<char, int> mp;
		int l = 0, r = s.size() - 1;
		int ans = 0;
		while (l < r)
		{
			if (!mp.count(s[l]))
			{
				ans++;
				mp[s[l]] = 1;
			}
			l++;
			r--;
		}
		if (ans > 1) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

B.

枚举多少个选 $1$ 操作。

假设为 $x$ 次。

那么就有 $k-x$ 次 $2$ 操作。

考虑 $x$ 次 $1$ 操作时，前 $2x$ 小的数会被删掉。$k-x$ 次 $2$ 操作时，最 $k-x$ 大的数会被删掉。

于是排序后前缀和即可。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <cstdlib>
#include <string>
using namespace std;

#define ll long long

const int N = 2e5 + 5, INF = 2e9, MOD = 1e9 + 7;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, t, a[N];
long long sum[N];

int main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]), sum[i] = 0;
		sort(a + 1, a + n + 1);
		for (int i = 1; i <= n; i++)
		{
			sum[i] = sum[i - 1] + a[i] * 1LL;
		}
		long long ans = 0LL;
		for (int i = 1; i <= n; i++)
		{
			ans += a[i];
		}
		long long minn = 1e18;
		for (int i = 0; i <= m; i++)
		{
			int c = m - i;
			minn = min(minn, sum[i * 2] + sum[n] - sum[n - c]);
		}
		ans -= minn;
		printf("%lld\n", ans);
	}
	return 0;
}

C.

容易发现对于一段单调递增或递减的 $a$，那么只需要取两端即可。

于是只需要求有多少个 $i$ 使得 $i$ 是递增和递减的转折点即可。

特判全部都相同即可。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <cstdlib>
#include <string>
using namespace std;

#define ll long long

const int N = 3e5 + 5, INF = 2e9, MOD = 1e9 + 7;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, t, a[N];

int main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
		int x = a[1];
		bool f = 1;
		for (int i = 2; i <= n; i++)
		{
			if (x != a[i])
			{
				f = 0;
				break;
			}
		}
		if (n == 1 || f)
		{
			printf("1\n");
			continue;
		}
		int ans = 1;
		int nowv, place;
		for (int i = 1; i < n; i++)
		{
			if (a[i] != a[i + 1])
			{
				nowv = a[i] < a[i + 1];
				place = i + 1;
				break;
			}
		}
		for (int i = place; i < n; i++)
		{
			if (a[i] == a[i + 1]) continue;
			if ((a[i] < a[i + 1]) != nowv)
			{
				ans++;
				nowv = (a[i] < a[i + 1]);
			}
		}
		printf("%d\n", ans + 1);
	}
	return 0;
}

D1.

二分。