Educational Codeforces Round 69 (Rated for Div. 2) A B C D

Educational Codeforces Round 69 (Rated for Div. 2)

A

#include
using namespace std;
typedef long long ll;
const int N = 1e5+100;

int t;
int n;
int a[N];
int main(){
	cin >> t;
	while(t--){
		cin >> n;
		for(int i=0;i> a[i];
		}
		sort(a,a+n);
		int h=min(a[n-1],a[n-2]);
		if(h<2){
			cout << "0\n";
		}
		else cout << min(n-2,h-1) << "\n";
	}
	return 0;
}

　　B

数组必须先增后减，或者单调递增，单调递减

#include
using namespace std;
typedef long long ll;
const int N = 2e5+100;
int n;
int a[N];
int main(){
	cin >> n;
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	int i;
	for(i = 1; i < n; i++){
		if(a[i] < a[i-1])
			break;
	}
	int flag = 1;
	for(i=i+1;i a[i-1]){
			flag = 0;
		}
	}
	if(flag) cout << "YES\n";
	else cout << "NO\n";
	return 0;
}

　　C 

要使数组分成k份，必须选择k-1个间隔，间隔差值越大，最后的每个区间max-min的和越小

#include
using namespace std;
typedef long long ll;
const int N = 3e5+100;

int n,k;
int a[N];
int p[N];
bool cmp(int x,int y){
	return x>y;
}
int main(){
	cin >> n >> k;
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	for(int i=1; i < n; i++){
		p[i] = a[i] - a[i-1];
	}
	sort(p+1,p+n,cmp);
	ll res = 0;
	for(int i=1; i<=k-1;i++){
		res += p[i];
	}
	cout << a[n-1]- a[0]-res << "\n";
	return 0;
}

　　D

#include
using namespace std;
typedef long long ll;
const int N=3e5+100;
int n,m,k;
int a[N];
ll sum[N];
ll dp[N];
int main(){
    cin >> n >> m >> k;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        dp[i] = max(0, a[i]-k);
        sum[i] = sum[i-1] + a[i];
    }
    ll res = 0;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(i-j >= 0){
                dp[i] = max(dp[i], dp[i-j]-k + sum[i]-sum[i-j]);
                res = max(res,dp[i]);
            }
        }
    }
    cout << res << endl;
}

　　

转载于:https://www.cnblogs.com/YJing814/p/11244151.html