CF1771

Hossam and Combinatorics

~~建议降红~~

相当于是最大值个数和最小值个数乘积再乘上 $2$

需要注意特判最大值和最小值相等的情况这时候的答案是 $n * (n - 1)$

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
#define getchar() cin.get()
#define int long long
const int inf = 0x3f3f3f3f;
const int N = 3e5 + 5;

int read()
{
	int f = 1 , x = 0;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , a[N] , maxx , minn , maxxcnt , minncnt;

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read();
	while ( T -- )
	{
		n = read();
		maxx = maxxcnt = minncnt = 0 , minn = inf;
		for ( int i = 1 ; i <= n ; i ++ ) a[i] = read() , maxx = max ( maxx , a[i] ) , minn = min ( minn , a[i] );
		for ( int i = 1 ; i <= n ; i ++ )
		{
			if ( maxx == a[i] ) ++ maxxcnt;
			if ( minn == a[i] ) ++ minncnt;
		}
		cout << ( maxx == minn ? ( n * ( n - 1 ) ) : maxxcnt * minncnt * 2 ) << endl;
	}
	return 0;
}

Hossam and Friends

我们设 $a [i]$ 表示 $a = i$ 时最大的 $b$

对于一个点对 $(x, y)$ 我们可以将 $a [x]$ 更新为 $m i n (a [x], y - 1)$

然后我们从后往前更新 $a [i] = m i n (a [i], a [i + 1])$ 这是因为上一个的最远点一定是这一个的最远点因为上一个的终点 $b$ 再大就不满足 $y_{i} > b$ 了那么这一个的终点也不应该比上一个的终点大(因为这一个的起点 $a$ 还要比上一个的起点 $a$ 小更有可能覆盖上一段点对)

注意 $x$ 有可能大于 $y$ 需要交换一下

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
#define getchar() cin.get()
#define int long long
const int N = 3e5 + 5;

int read()
{
	int f = 1 , x = 0;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , a[N] , m;

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read();
	while ( T -- )
	{
		n = read() , m = read();
		fill ( a + 1 , a + n + 1 , n );
		for ( int i = 1 , x , y ; i <= m ; i ++ ) x = read() , y = read() , a[min(x,y)] = min ( a[min(x,y)] , max(x,y) - 1 );
		for ( int i = n - 1 ; i ; i -- ) a[i] = min ( a[i] , a[i+1] );
		int ans = 0;
		for ( int i = 1 ; i <= n ; i ++ ) ans += ( a[i] - i + 1 );
		cout << ans << endl;
	}
	return 0;
}

Hossam and Trainees

明显比上一场简单...

我们可以先筛出 $\sqrt{n} (n = 1 e 9)$ 之间的所有质数开一个 $c n t$ 数组记录每一个质数作为因数的次数

我们对于一个数将它质因数分解并记录在 $c n t$ 数组中

对于 $> \sqrt{n}$ 的因数来说因为最多这样的因数只有一个那么我们可以用 $m a p$ 来记录这个因数是否被访问过了

最后我们遍历 $c n t$ 数组并统计是否有被统计超过两次的质因数输出答案即可

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
#define getchar() cin.get()
const int N = 1e5 + 5;
const int maxn = sqrt(1e9);

int read()
{
	int f = 1 , x = 0;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , a[N] , ans , cnt[N];
map<int,int> mp;

int isntprime[N] , prime[N] , tot;
void pre ()
{
	for ( int i = 2 ; i <= maxn ; i ++ )
	{
		if ( !isntprime[i] ) prime[++tot] = i;
		for ( int j = 1 ; j <= tot && i * prime[j] <= maxn ; j ++ )
		{
			isntprime[i*prime[j]] = 1;
			if ( ! ( i % prime[j] ) ) break;
		}
	}
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read();
	pre();
	while ( T -- )
	{
		ans = 0 , mp.clear();
		memset ( cnt , 0 , sizeof cnt );
		n = read();
		for ( int i = 1 ; i <= n ; i ++ ) a[i] = read();
		for ( int i = 1 ; i <= n ; i ++ )
		{
			for ( int j = 1 ; j <= tot && a[i] >= prime[j] ; j ++ )
				if ( ! ( a[i] % prime[j] ) )
				{
					++ cnt[j];
					while ( ! ( a[i] % prime[j] ) ) a[i] /= prime[j];
				}
			if ( a[i] > 1 )
			{
				if ( mp[a[i]] ) { ans = 1; break; }
				mp[a[i]] = 1;
			}
		}
		for ( int i = 1 ; i <= tot ; i ++ ) ans |= ( cnt[i] >= 2 );
		cout << ( ans ? "YES" : "NO" ) << endl;
	}
	return 0;
}

Hossam and (sub-)palindromic tree

难度起来了()

先考虑序列上的情况设 $f (l, r)$ 表示 $s_{l \dots r}$ 的最长回文子序列长度可得到转移方程：

f (l, r) = max {f (l + 1, r), f (l, r - 1), f (l + 1, r - 1) + 2 [s_{l} = s_{r}]}

将这个东西转移到树上即设 $f (x, y)$ 表示 $x, y$ 路径构成的最长回文子序列长度 $f a (x, y)$ 表示以 $x$ 为根时 $y$ 的父亲可得转移方程：

f (x, y) = max {f (f a (y, x), y), f (x, f a (x, y)), f (f a (y, x), f a (x, y)) + 2 [s_{x} = s_{y}]}

最终答案即为 $max_{x, y} f (x, y)$ 记忆化搜索实现即可

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
#define getchar() cin.get()
const int N = 2e3 + 5;

int read()
{
	int f = 1 , x = 0;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , f[N][N] , fa[N][N] , dep[N][N] , ans;
string s;

vector<int> e[N];
inl void add ( int u , int v ) { e[u].eb(v); }

void clear()
{
	for ( int i = 1 ; i <= n ; i ++ ) for ( int j = 1 ; j <= n ; j ++ ) fa[i][j] = dep[i][j] = 0 , f[i][j] = -1;
	for ( int i = 1 ; i <= n ; i ++ ) e[i].clear();
	ans = 0;
}

void dfs ( int rt , int u , int ff )
{
	fa[rt][u] = ff , dep[rt][u] = dep[rt][ff] + 1;
	for ( auto v : e[u] ) if ( v ^ ff ) dfs ( rt , v , u );
}

int dp ( int x , int y )
{
	if ( f[x][y] != -1 ) return f[x][y];
	if ( dep[x][y] == 1 ) return f[x][y] = 1;
	if ( dep[x][y] == 2 ) return f[x][y] = ( s[x] == s[y] ) + 1;
	return f[x][y] = max ( max ( dp ( fa[y][x] , y ) , dp ( x , fa[x][y] ) ) , dp ( fa[y][x] , fa[x][y] ) + ( ( s[x] == s[y] ) ? 2 : 0 ) );
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read();
	while ( T -- )
	{
		n = read();
		clear();
		cin >> s; s = " " + s;
		for ( int i = 1 , u , v ; i < n ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
		for ( int i = 1 ; i <= n ; i ++ ) dfs ( i , i , 0 );
		for ( int i = 1 ; i <= n ; i ++ ) for ( int j = i ; j <= n ; j ++ ) ans = max ( ans , dp ( i , j ) );
		cout << ans << endl;
	}
	return 0;
}

Hossam and a Letter

类似种花的题

记录 $u p [i] [j] [0 / 1]$ 表示从 $(i, j)$ 点向上走经过 $0 / 1$ 个 $m$ 能到达的最远点 $d o w n [i] [j] [0 / 1]$ 同理

那么我们 $O (n^{3})$ 地枚举每一条横杠进行统计即可(具体来说枚举横杠的左端点然后再枚举右端点的纵坐标即可)

注意当 $(i, j)$ 本身是 $m$ 点的时候要提前处理使得 $u p [i] [j] [0] = u p [i] [j] [1]$ $d o w n$ 也同理这是为了在查询的时候不受到枚举的横杠左端点是 $m$ 点的干扰

细节很多状态转移方程不要挂了

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define mid (l+(r-l)/2)
// #define mid ((l+r)/2)
#define ls(p) t[p].son[0]
#define rs(p) t[p].son[1]
#define lson ls(p),l,mid
#define rson rs(p),mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define getchar() cin.get()
// #define int long long
const int N = 1e3 + 5;
const int maxn = 2e9; 

int read()
{
	int f = 1 , x = 0;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , up[N][N][2] , dw[N][N][2] , ans;

char mp[N][N];

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 ; i <= n ; i ++ )
		for ( int j = 1 ; j <= m ; j ++ )
			cin >> mp[i][j];
	for ( int i = 1 ; i <= n ; i ++ )
		for ( int j = 1 ; j <= m ; j ++ ) 
		{
			if ( mp[i][j] == '#' ) up[i][j][0] = up[i][j][1] = 0;
			else if ( mp[i][j] == '.' ) up[i][j][0] = up[i-1][j][0] + 1 , up[i][j][1] = up[i-1][j][1] + 1;
			else up[i][j][0] = 0 , up[i][j][1] = up[i-1][j][0] + 1;
		}
	for ( int i = n ; i ; i -- )
		for ( int j = 1 ; j <= m ; j ++ ) 
		{
			if ( mp[i][j] == '#' ) dw[i][j][0] = dw[i][j][1] = 0;
			else if ( mp[i][j] == '.' ) dw[i][j][0] = dw[i+1][j][0] + 1 , dw[i][j][1] = dw[i+1][j][1] + 1;
			else dw[i][j][0] = 0 , dw[i][j][1] = dw[i+1][j][0] + 1;
		}
	for ( int i = 1 ; i <= n ; i ++ )
		for ( int j = 1 ; j <= m ; j ++ ) 
			if ( mp[i][j] == 'm' ) up[i][j][0] = up[i][j][1] , dw[i][j][0] = dw[i][j][1];
	for ( int i = 2 ; i < n ; i ++ )
		for ( int j = 1 ; j <= m - 2 ; j ++ )//横杠左端点(i,j)
		{
			int cnt = ( mp[i][j] == 'm' ) + ( mp[i][j+1] == 'm' );
			if ( cnt > 1 || mp[i][j] == '#' || mp[i][j+1] == '#' ) continue;
			for ( int k = j + 2 ; k <= m ; k ++ )//横杠右端点(i,k)
			{
				if ( mp[i][k] == '#' ) break;
				cnt += ( mp[i][k] == 'm' );
				if ( cnt > 1 ) break;
				int res = 0;
				if ( up[i][j][0] > 1 && up[i][k][0] > 1 && dw[i][j][0] > 1 && dw[i][k][0] > 1 ) res = max ( res , k - j - 3 + min ( up[i][j][0] , up[i][k][0] ) * 2 + min ( dw[i][j][0] , dw[i][k][0] ) * 2 );
				if ( !cnt ) 
				{
					if ( up[i][j][1] > 1 && up[i][k][0] > 1 && dw[i][j][0] > 1 && dw[i][k][0] > 1 ) res = max ( res , k - j - 3 + min ( up[i][j][1] , up[i][k][0] ) * 2 + min ( dw[i][j][0] , dw[i][k][0] ) * 2 );
					if ( up[i][j][0] > 1 && up[i][k][1] > 1 && dw[i][j][0] > 1 && dw[i][k][0] > 1 ) res = max ( res , k - j - 3 + min ( up[i][j][0] , up[i][k][1] ) * 2 + min ( dw[i][j][0] , dw[i][k][0] ) * 2 );
					if ( up[i][j][0] > 1 && up[i][k][0] > 1 && dw[i][j][1] > 1 && dw[i][k][0] > 1 ) res = max ( res , k - j - 3 + min ( up[i][j][0] , up[i][k][0] ) * 2 + min ( dw[i][j][1] , dw[i][k][0] ) * 2 );
					if ( up[i][j][0] > 1 && up[i][k][0] > 1 && dw[i][j][0] > 1 && dw[i][k][1] > 1 ) res = max ( res , k - j - 3 + min ( up[i][j][0] , up[i][k][0] ) * 2 + min ( dw[i][j][0] , dw[i][k][1] ) * 2 );
				}
				ans = max ( ans , res );
			}
		}
	cout << ans << endl;
	return 0;
}

Hossam and Range Minimum Query

一个粗略的想法是我们用主席树维护异或前缀和因为异或的优秀性质我们可以得知如果区间内该数的出现次数是偶数那么整体异或值为 $0$

但这样做显然是错的例如 $(1100)_{2} \oplus (0011)_{2} = (1111)_{2}$ 这三个数都只出现了一次但是异或起来为 $0$

我们考虑为每一个值赋一个随机哈希值显然这样出现错误的概率会小很多用主席树维护即可

注意这里我们如果不开 $l o n g l o n g$ 的话对于 $m i d$ 的 $d e f i n e$ 就需要采用 $(l + (r - l) / 2)$ 的形式来防止溢出 $i n t$

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define mid (l+(r-l)/2)
// #define mid ((l+r)/2)
#define ls(p) t[p].son[0]
#define rs(p) t[p].son[1]
#define lson ls(p),l,mid
#define rson rs(p),mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define getchar() cin.get()
// #define int long long
const int N = 2e5 + 5;
const int maxn = 2e9; 

int read()
{
	int f = 1 , x = 0;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , q , a[N] , root[N] , lstans;
string s;
map<int,int> mp;

struct Tree
{
	struct node { int son[2] , sum; } t[N<<5];
	int tot = 0;
	inl int new_node ( int x ) { t[++tot] = t[x]; return tot; }
	void upd ( int &p , int l , int r , int x , int val )
	{
		p = new_node(p) , t[p].sum ^= val;
		if ( l == r ) return;
		if ( x <= mid ) upd ( lson , x , val );
		else upd ( rson , x , val );
	}
	int query ( int u , int v , int l , int r )
	{
		if ( l == r ) return l;
		if ( t[ls(u)].sum ^ t[ls(v)].sum ) return query ( ls(u) , ls(v) , l , mid );
		else return query ( rs(u) , rs(v) , mid + 1 , r );
	}
}T;

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	mt19937 rnd(random_device{}());
	n = read();
	for ( int i = 1 ; i <= n ; i ++ )
	{
		a[i] = read();
		if ( !mp.count(a[i]) ) mp[a[i]] = rnd();
		T.upd ( root[i] = root[i-1] , 0 , maxn , a[i] , mp[a[i]] );
	}
	q = read();
	for ( int i = 1 ; i <= q ; i ++ )
	{
		int l = read() ^ lstans , r = read() ^ lstans;
		lstans = T.query ( root[l-1] , root[r] , 0 , maxn );
		if ( lstans == maxn ) lstans = 0;
		cout << lstans << endl;
	}
	return 0;
}

posted @ 2023-10-25 18:23 Echo_Long 阅读(490) 评论(0) 编辑收藏举报

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Echo-Long

CF1771

CF1771

Hossam and Combinatorics

Hossam and Friends

Hossam and Trainees

Hossam and (sub-)palindromic tree

Hossam and a Letter

Hossam and Range Minimum Query

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