Codeforces Round #821 (Div. 2)A~D1

A. Consecutive Sum

int n,k; 
const int N=110;
int a[N];
void solve()
{
	memset(a,0,sizeof a);
	cin>>n>>k;
	rep(i,1,n)cin>>a[i];
	vector<int>v;
	rep(i,1,k)
	{
		int maxv=0;
		for(int j=i;j<=n;j+=k)maxv=max(maxv,a[j]);
		v.pb(maxv);
	}
	int s=0;
	sort(all(v),greater<int>());
	rep(i,0,k-1)s+=v[i];
	cout<<s<<endl;
}

B. Rule of League

int n,a,b;
void solve()
{
	cin>>n>>a>>b;
	if(a&&b||!a&&!b)
	{
		cout<<-1<<endl;
		return;
	}
	else
	{
		if((n-1)%(a+b))cout<<-1<<endl;
		else
		{
			int s=a+b;
			int res=2;
			for(int i=1;i<n;i+=s)
			{
				for(int k=1;k<=s;k++)cout<<res<<' ';
				res+=s;
			}
			cout<<endl;
		}
	}
}

C. Parity Shuffle Sorting

int n;
const int N=1e5+10;
int a[N];
void solve()
{
	cin>>n;
	rep(i,1,n)cin>>a[i];
	bool f=false;
	rep(i,1,n-1)if(a[i]>a[i+1])f=true;
	if(!f)
	{
		cout<<0<<endl;
		return;
	}
	int res=0;
	vector<PII>v;
	if((a[1]+a[n])%2)a[n]=a[1],res++,v.pb({1,n});
	else a[1]=a[n],res++,v.pb({1,n});
	rep(i,2,n-1)
	{
		if((a[i]+a[1])%2)a[i]=a[1],res++,v.pb({1,i});
		else res++,v.pb({i,n});
	}
	cout<<res<<endl;
	for(auto x:v)
		cout<<x.x<<' '<<x.y<<endl;
}

D1. Zero-One (Easy Version)

#include 
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'//交互题删掉此 
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){return (a%mod+mod)%mod;}
int lowbit(int x){return x&-x;}//最低位1及其后面的0构成的数值
int qmi(int a, int k, int p){int res = 1 % p;while (k){if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
int inv(int a,int mod){return qmi(a,mod-2,mod);}
int n,x,y;
string a,b;
void solve()
{
	cin>>n>>x>>y;
	cin>>a>>b;
	if(a==b)
	{
		cout<<0<<endl;
		return;
	}
	bool f=false;
	int cnt=0;
	if(x>=y*2)f=true;
	vector<int>v;
	for(int i=0;i<n;i++)
		if(a[i]!=b[i])v.pb(i),cnt++;
	if(v.size()%2==0)
	{
		if(v.size()==2&&v[0]+1==v[1])
		{
			if(f)cout<<y*2<<endl;
			else cout<<x<<endl;
		}
		else cout<<cnt/2*y<<endl;
	}
	else cout<<-1<<endl;
}
signed main()
{
	io;
	int _;_=1;
	cin>>_;
	while(_--)solve();
    return 0;
}