A

#include <bits/stdc++.h>
 
using namespace std;
typedef long long LL;
const int N = 60;
char c[N];
 
void run()
{	
	scanf("%s", c + 1);
	int n = strlen(c + 1);
	map<char, int>st;
	st[c[1]] ++;
	for(int i = 2; i <= n / 2; ++ i)
	{
		if(st.find(c[i]) == st.end())
		{
			printf("YES\n");
			return;	
		}
	}
	printf("NO\n");
}
 
int main()
{
//	freopen("1.in", "r", stdin);
	int t;cin >> t;
	while(t --)	run();	
	return 0;  
}

B

直接枚举即可

#include <bits/stdc++.h>
 
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
int a[N];
LL s[N];
 
void run()
{	
	int n, k;
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; ++ i)	scanf("%d", &a[i]);
	sort(a + 1, a + 1 + n);	
	for(int i = 1; i <= n; ++ i)	s[i] = s[i - 1] + a[i];
	LL ans = 0;
	for(int i = 0; i <= k; ++ i)
	{
		int l = 2 * (k - i), r = n - i;
		ans = max(ans, s[r] - s[l]);
	}
	printf("%lld\n", ans);
}
 
int main()
{
//	freopen("1.in", "r", stdin);
	int t;cin >> t;
	while(t --)	run();	
	return 0;  
}

C

C比赛时没出，思路上没什么大问题，就是差一点（bushi，说到底还是自己菜）
我们可以发现这道题只需要找一下有多少个波峰和波谷就可以了，因为波峰波谷一定不能删除

#include <bits/stdc++.h>
 
using namespace std;
typedef long long LL;
const int N = 3e5 + 10;
 
void run()
{	
	int n;
	scanf("%d", &n);
	vector<int>a(n);
	for(int i = 0; i < n; ++ i)		scanf("%d", &a[i]); 
	n = unique(a.begin(), a.end()) - a.begin(); 
	int ans  = 0;
	for(int i = 0; i < n; ++ i)
	{
		if(i == 0 || i == n -1 || a[i] > a[i - 1] == a[i] > a[i + 1])	ans ++;
	}
	printf("%d\n", ans);
}
 
int main()
{
//	freopen("1.in", "r", stdin);
	int t;cin >> t;
	while(t --)	run();	
	return 0;  
}