Codeforces Round #573 (Div. 2)

最新推荐文章于 2020-02-04 23:06:37 发布

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最新推荐文章于 2020-02-04 23:06:37 发布

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分类专栏： acm 文章标签： codeforces

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Sloved	A	B	C	D	E	F
6/6	O	O	O	O	Ø	Ø

O for passing during the contest
Ø for passing after the contest
! for attempted but failed
· for having not attempted yet

A.Tokitsukaze and Enhancement(0:05,+1)

签到题，读题找规律即可

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 
int n;
int main(int argc, char const *argv[])
{
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	int x = n % 4;
	if(x == 1)
	{
		cout << 0  << " " << "A"<< endl;
	}
	if(x == 2)
	{
		cout << 1  << " " << "B"<< endl;
	}
	if(x == 3)
	{
		cout << 2  << " " << "A"<< endl;
	}
	if(x == 0)
	{
		cout << 1  << " " << "A"<< endl;
	}
	return 0;
}

B.Tokitsukaze and Mahjong(0:24,+3)

模拟，注意1m 3m 5m的情况

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 
string str[10];
int cnt[5][10];
int main(int argc, char const *argv[])
{
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	rep(i,0,3)
	{
		cin >> str[i];
		if(str[i][1] == 's')
		{
			cnt[0][str[i][0]-'0']++;
		}
		else if(str[i][1] == 'p')
		{
			cnt[1][str[i][0]-'0']++;
		}
		else
		{
			cnt[2][str[i][0]-'0']++;
		}
	}
	int ans = 2;
	//
	rep(i,0,3)
	{
		rep(j,1,10)
		{
			if(cnt[i][j])
			{
				ans = min(ans,3-cnt[i][j]);
			}
		}
	}
	int s = 0;
	rep(i,0,3)
	{
		s = 0;
		rep(j,1,10)
		{
			if(cnt[i][j])
			{
				s+=1;
			}
			else
			{
				ans = min(ans,3-s);
				s = 0;
			}
		}
		ans = min(ans,3-s);
	}
	rep(i,0,3)
	{
		rep(j,1,8)
		{
			if(cnt[i][j] && cnt[i][j+2])
			{
				ans = min(ans,1);
			}
		}
	}
	cout << ans << endl;
	return 0;
}

C.Tokitsukaze and Discard Items(00:48,+1)

直接O(n)遍历模拟即可，不要想复杂

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxn = 1e5+100;
 
 
ll n,m,k;
ll p;
int main(int argc, char const *argv[])
{
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m >> k;
	ll ans = 0;
	ll t = 0;//dz
	ll s = 0;//-
	ll s1 = 0;
	rep(i,0,m)
	{
		cin >> p;
		p -= s;
		if((p%k == 0 && p / k == t)||(p%k != 0 && p / k + 1 == t))
		{
			s1++;
		}
		else
		{
			p -= s1 - s;
			s = s1;
			ans++;
			//cout << i << "----" << endl;
			if(p % k == 0)
			{
				t = p / k;
			}
			else
				t = p / k + 1;
			s1++;
		}
	}
	cout << ans << endl;
	return 0;
}

D.Tokitsukaze, CSL and Stone Game(01:35,+2)

考虑几种情况

2 2 3和1 2 2，后一种cslnb
计算数的个数大于2的总数，cnt >= 2，则cslnb（包括 3 3 3这种情况）
一些显然得到答案的方案

最后特判完，假如还没有得到结果，则最后的局面应为 0 1 2 3 … n-1，直接作差，%2判断即可

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxn = 1e5 + 1000;
 
ll n;
ll a[maxn];
int main(int argc, char const *argv[])
{
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
 
	cin >> n;
 
	ll sum = 0;
	int zero = 0;
	rep(i,0,n)
	{
		cin >> a[i];
		if(a[i] == 0)
		{
			zero++;
		}
		sum += a[i];
	}
	if(sum == 0)
	{
		cout << "cslnb" << endl;
		return 0;
	}
 
	if(zero >= 2)
	{
		cout << "cslnb" << endl;
		return 0;	
	}
 
	if(n == 1) // judge only
	{
		if(sum % 2 == 0)
		{
			cout << "cslnb"<<endl;
		}
		else
		{
			cout << "sjfnb"<<endl;
		}
		return 0;
	}
 
	sort(a,a+n);
 
	int cnt = 0;
	int t = 0;
	int wz = 0;
	rep(i,0,n)
	{
		if(a[i] == a[i+1])
		{
			cnt++;
			t = a[i];
			wz = i;
		}
	}
	if(cnt >= 2)
	{
		   cout <<"cslnb" << endl;
		   return 0;
	}
	else if(cnt == 1)
	{
		if(wz == 0 || a[wz-1] < t-1)
		{
			sum -= (0+n-1)*n/2;
			if(sum % 2 == 0)
			{
				cout << "cslnb"<<endl;
			}
			else
			{
				cout << "sjfnb"<<endl;
			}
			return 0;
		}
		else
		{
			cout << "cslnb" << endl;
			return 0;
		}
	}
	else
	{
		sum -= (0+n-1)*n/2;
		if(sum % 2 == 0)
		{
			cout << "cslnb"<<endl;
		}
		else
		{
			cout << "sjfnb"<<endl;
		}
	}
	return 0;

E.Tokitsukaze and Duel

找出第一个和最后一个1，第一个和最后一个0
sjf胜利的条件很简单，zero2 - zero1 < k || one2- one1 < k
难点在qls的回合判断
现在设计一个数组a，元素为zero2 ， zero1 ， one2， one1，并排序

已知在qls的回合，qls不可能输，顶多once again

qls胜利，条件为a2-a0 <= k and a3-a1 <= k

已经判断了第一回合，则a1和a2不可能同是0或1，即a0和a3为0和1
无论sjf选择哪一段，必包括a1和a2
假设sjf把那一段变成1，qls可以把边缘的0到中间一段变成1，同理假设sjf把那一段变成0，qls可以把边缘的1到中间一段变成0
重点在于无论如何，当满足条件时，sjf无论如何操作都覆盖到了中间的0和1

假设不满足这个条件，则sjf必定可以选择一段不覆盖中间的0和1，qls只能选择once again
具体要理解清楚可能需要自己举一些例子，或者严格证明一下

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

int n,k;
string str;
int one1,one2;
int zero1,zero2;

int a[10];
int main(int argc, char const *argv[])
{
	cin >> n >> k;
	cin >> str;

	rep(i,0,n)
	{
		if(str[i] == '0')
		{
			if(zero1 == 0)
			{
				zero1 = i + 1;
			}
				zero2 = i + 1;
		}
		if(str[i] == '1')
		{
			if(one1 == 0)
			{
				one1 = i + 1;
			}
				one2 = i + 1;
		}
	}
	if(zero2 - zero1 < k || one2 - one1 < k)
	{
		cout << "tokitsukaze" << endl;
		return 0;
	}
	a[0] = zero1;
	a[1] = zero2;
	a[2] = one1;
	a[3] = one2;
	sort(a,a+4);
	if(a[3] - a[1] <= k && a[2] - a[0] <= k)
	{
		cout << "quailty" << endl;
		return 0;
	}
	cout << "once again" << endl;
	return 0;
}

F.Tokitsukaze and Strange Rectangle

扫描线+离散化

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//树状数组
struct Bit
{
    vector<int> a;
    int sz;
    void init(int n)
    {
        sz=n+5;
        for(int i=1;i<=n+5;i++)
            a.push_back(0);
    }
    int lowbit(int x)
    {
        return x&(-x);
    }
    int query(int x)
    {
        int ans = 0;
        for(;x;x-=lowbit(x))ans+=a[x];
        return ans;
    }
    int query(int l ,int r)
    {
    	return query(r) - query(l-1);
    }
    void update(int x,int v)
    {
        for(;x<sz;x+=lowbit(x))
            a[x]+=v;
    }
}bit;
 
const int maxn = 2e5 + 100;
VI x[maxn],y; // lsy x, unique y
 
map<int,int> lsx,lsy,cntx;// id of discrete x,y  , number of x
 
int n;
int main(int argc, char const *argv[])
{
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	bit.init(n);
 
 
	int sum = 0;
	rep(i,0,n)
	{
		int xx,yy;
		cin >> xx >> yy;
		if(!lsy.count(yy))
		{
			lsy[yy] = sum++;
			y.pb(yy);
		}
		x[lsy[yy]].pb(xx);
		cntx[xx]++;
	}
	sort(all(y));
	// for(auto v : y)
	// {
	// 	sort(all(x[lsy[v]]));
	// }
 
	int s = 0;
	for(auto v:cntx)
	{
		s++;
		lsx[v.fi] = s;
		bit.update(s,1);
	}
	
	ll ans = 0;
	for(auto v : y)
	{
		sort(all(x[lsy[v]]));		
		int pre = 1; 
		for(auto u : x[lsy[v]])
		{
			ans += 1ll * bit.query(pre,lsx[u]) * bit.query(lsx[u],s);
			//cout << pre << " "<< lsx[u] <<" " << bit.query(pre,lsx[u]) <<  "---" << endl;
			//cout << s << " " << bit.query(lsx[u],s) << endl;
			pre = lsx[u] + 1;
			cntx[u]--;
			if(cntx[u] == 0)
			{
				//cntx.erase(u);
				bit.update(lsx[u],-1);
			}													
		}
	}
	cout << ans << endl;
	return 0;
}