[考试记录] 2024.9.16 csp-s模拟赛30

1.[考试记录] 2024.5.192024-05-21 2.[考试记录] 2024.6.92024-06-09 3.[考试记录] 2024.7.52024-07-05 4.[考试记录] 2024.7.15 csp-s模拟赛42024-07-15 5.[考试记录] 2024.8.10 csp-s 模拟赛182024-08-10 6.[考试记录] 2024.8.14 csp-s模拟赛202024-08-14 7.[考试记录] 2024.8.17 csp-s模拟赛212024-08-17 8.[考试记录] 2024.9.22 csp-s模拟赛312024-09-22 9.[考试记录] 2027.9.15 csp-s 模拟赛292024-09-22

10.[考试记录] 2024.9.16 csp-s模拟赛302024-09-22

11.[考试记录] 2024.10.7 csp-s模拟赛372024-10-08 12.[考试记录] 2024.11.7 noip模拟赛72024-11-07 13.[考试记录] 2024.11.9 noip模拟赛92024-11-10 14.[考试记录] 2024.11.12 noip模拟赛112024-11-12 15.[考试记录] 2024.11.19 noip模拟赛172024-11-19 16.[考试记录] 2024.11.22 noip模拟赛192024-11-22

T1 不相邻集合

服了，考场上拉一泡权值线段树，硬是没调过来。然后一下考就知道是哪的问题了，服了~

维护两个东西：一是以 $x$ 为右端点的最长可重集的长度，二是以 $y$ 为右端点的最长可重集长度。两者一加再减一就是答案。然后考场上愣是想不到。这玩意用权值线段树很好维护，考虑到因为是以 $x$ 为右端点的最长长度就是把以 $1\sim x-2$ 中的最长长度加一即可。

 #include<bits/stdc++.h>
using namespace std;
constexpr int B = 1 << 25;
char buf[B], *p1 = buf, *p2 = buf;
#define gt() (p1==p2 && (p2=(p1=buf)+fread(buf, 1, B, stdin), p1==p2) ? EOF : *p1++)
template <typename T> inline void rd(T &x){
	x = 0; int f = 0; char ch = gt();
	for(; !isdigit(ch); ch = gt()) f ^= ch == '-';
	for(; isdigit(ch); ch = gt()) x = (x<<1) + (x<<3) + (ch^48);
	x = f ? -x : x;
}
char obuf[B], *O = obuf;
#define pt(ch) (O-obuf==B && (fwrite(obuf, 1, B, stdout), O=obuf), *O++ = (ch))
template <typename T> inline void wt(T x){
	if(x < 0) pt('-'), x = -x;
	if(x >= 10) wt(x / 10); pt(x % 10 ^ 48);
}
#define fw fwrite(obuf, 1, O - obuf, stdout)
constexpr int N = 3e5 + 5, T = 1e6 + 5, M = 5e5 + 1;
int n, x, ans, mx, rt, as1, as2;
namespace ST{
	int cnt, ls[T], rs[T], f[T], g[T];
	inline void fadd(int &id, int l, int r, int x, int val){
		if(!id) id = ++cnt;
		if(l == r){
			f[id] = max(f[id], val);
			return;
		}
		int mid = (l + r) >> 1;
		if(x <= mid) fadd(ls[id], l, mid, x, val);
		else fadd(rs[id], mid+1, r, x, val);
		f[id] = max(f[ls[id]], f[rs[id]]);
	}
	inline void gadd(int &id, int l, int r, int x, int val){
		if(!id) id = ++cnt;
		if(l == r){
			g[id] = max(g[id], val);
			return;
		}
		int mid = (l + r) >> 1;
		if(x <= mid) gadd(ls[id], l, mid, x, val);
		else gadd(rs[id], mid+1, r, x, val);
		g[id] = max(g[ls[id]], g[rs[id]]);
	}
	inline int fquery(int &id, int l, int r, int x, int y){
		if(!id || x > y) return 0;
		if(x <= l && r <= y) return f[id];
		int mid = (l + r) >> 1, ans = 0;
		if(x <= mid) ans = max(ans, fquery(ls[id], l, mid, x, y));
		if(y >  mid) ans = max(ans, fquery(rs[id], mid+1, r, x, y));
		return ans;
	}
	inline int gquery(int &id, int l, int r, int x, int y){
		if(!id || x > y) return 0;
		if(x <= l && r <= y) return g[id];
		int mid = (l + r) >> 1, ans = 0;
		if(x <= mid) ans = max(ans, gquery(ls[id], l, mid, x, y));
		if(y >  mid) ans = max(ans, gquery(rs[id], mid+1, r, x, y));
		return ans;
	}
}
bitset<M> vis;
int main(){
	rd(n); while(n--){
		rd(x); if(vis[x]){ wt(ans), pt(' '); continue; }
		vis[x] = 1;
		as1 = ST::fquery(rt, 1, M, 1, x-2) + 1;
		ST::fadd(rt, 1, M, x, as1);
		as2 = ST::gquery(rt, 1, M, x+2, M) + 1;
		ST::gadd(rt, 1, M, x, as2);
		ans = max(ans, as1 + as2 - 1);
		wt(ans), pt(' ');
	} return fw, 0;
}

T2 线段树

solution1

对于 $n\le 300$ 的部分，直接建树模拟即可。

solution 2

对于 $n=2^k$ 的部分。考虑遍历树的每一层，发现每一层树上节点的编号是有序的，那么找出 $x$ 和 $y$ 包含的完整段编号即可用等差数列求和计算出贡献。对于第 $i$ 层，起始编号为 $a = 2^{i-1}$ ，块长为 $b = 2^{\log_2(n)-i+1}$ 。那么 $x$ 完全包含的块的编号为：

l = {\begin{matrix} a + ⌊ \frac{x - 1}{b} ⌋ & b ∣ x - 1 \\ a + ⌊ \frac{x - 1}{b} ⌋ + 1 & b ∤ x - 1 \end{matrix}

$l=\left\{\begin{matrix} a+\left \lfloor \frac{x-1}{b} \right \rfloor & b\mid x-1\\ a+\left \lfloor \frac{x-1}{b} \right \rfloor+1 & b\nmid x-1 \end{matrix}\right.$

同理 $y$ 完全包含的块的编号为：

r = {\begin{matrix} a + ⌊ \frac{y - 1}{b} ⌋ & b ∣ y \\ a + ⌊ \frac{y - 1}{b} ⌋ - 1 & b ∤ y \end{matrix}

$r=\left\{\begin{matrix} a+\left \lfloor \frac{y-1}{b} \right \rfloor & b\mid y\\ a+\left \lfloor \frac{y-1}{b} \right \rfloor-1 & b\nmid y\end{matrix}\right.$

每一层完全包含的块构成等差数列，带公式即可。

solution 3

对于 $x=1,y=n$ 的部分，机房大佬打表得出答案为 $\binom{n}{2}$ 。但是由于 $n$ 极大，实际上不好打（？打不了）。

正解

令 $f(n,x)$ 表示有 $n$ 个叶子，根节点编号为 $x$ 的线段树的答案。显然有 $f(n,x)=k_nx+b_n$ 。很好理解：当根节点的编号加上 $p$ 时，儿子节点加上 $2p$ ，孙子加上 $4p$ ……显然一次函数， $k_n$ 只和树的形态有关。又因为：

f (n, x) = f (⌈ \frac{n}{2} ⌉, 2 x) + f (⌊ \frac{n}{2} ⌋, 2 x + 1) + x

$f(n,x)=f(\left \lceil \frac{n}{2} \right \rceil,2x)+f(\left \lfloor \frac{n}{2} \right \rfloor ,2x+1)+x$

那就暴力拆狮子：

{\begin{matrix} k_{n} = 2 k_{⌈ \frac{n}{2} ⌉} + 2 k ⌊ \frac{n}{2} ⌋ + 1 \\ b_{n} = b_{⌈ \frac{n}{2} ⌉} + b_{⌊ \frac{n}{2} ⌋} + k_{⌊ \frac{n}{2} ⌋} \end{matrix}

$\left\{\begin{matrix} k_n=2k_{\left \lceil \frac{n}{2} \right \rceil} +2k{\left \lfloor \frac{n}{2} \right \rfloor}+1\\ b_n=b_{\left \lceil \frac{n}{2} \right \rceil}+b_{\left \lfloor \frac{n}{2} \right \rfloor}+k_{\left \lfloor \frac{n}{2} \right \rfloor} \end{matrix}\right.$

记搜即可。

 #include<bits/stdc++.h>
using namespace std;
constexpr int B = 1 << 25;
char buf[B], *p1 = buf, *p2 = buf;
#define gt() (p1==p2 && (p2=(p1=buf)+fread(buf, 1, B, stdin), p1==p2) ? EOF : *p1++)
template <typename T> inline void rd(T &x){
	x = 0; int f = 0; char ch = gt();
	for(; !isdigit(ch); ch = gt()) f ^= ch == '-';
	for(; isdigit(ch); ch = gt()) x = (x<<1) + (x<<3) + (ch^48);
	x = f ? -x : x;
}
char obuf[B], *O = obuf;
#define pt(ch) (O-obuf==B && (fwrite(obuf, 1, B, stdout), O=obuf), *O++ = (ch))
template <typename T> inline void wt(T x){
	if(x < 0) pt('-'), x = -x;
	if(x >= 10) wt(x / 10); pt(x % 10 ^ 48);
}
#define fw fwrite(obuf, 1, O - obuf, stdout)
#define it __int128
#define int long long
#define ls (id << 1)
#define rs (id << 1 | 1)
constexpr int M = 1e9 + 7;
int T, n, x, y;
unordered_map<int, int> k, b;
inline int getk(int num){
	if(k.count(num)) return k[num];
	return k[num] = ((it)getk(num>>1) * 2 % M + (it)getk((num+1)>>1) * 2 % M + 1) % M;
}
inline int getb(int num){
	if(b.count(num)) return b[num];
	return b[num] = ((it)getb(num>>1) + (it)getk(num>>1) + (it)getb((num+1)>>1)) % M;
}
inline int work(int id, int num){
	return ((it)getk(num) * id % M + (it)getb(num)) % M;
}
inline int getans(int id, int l, int r, int x, int y){
	if(x <= l && r <= y) return work(id, r-l+1);
	int mid = (l + r) >> 1, ans = 0;
	if(x <= mid) ans = ((it)ans + (it)getans(ls, l, mid, x, y)) % M;
	if(y >  mid) ans = ((it)ans + (it)getans(rs, mid+1, r, x, y)) % M;
	return ans;
}
signed main(){
	k[1] = 1, b[1] = 0;
	rd(T); while(T--){
		rd(n), rd(x), rd(y);
		wt(getans(1, 1, n, x, y)); pt('\n');
	} return fw, 0;
}

T3 魔法师

昨天刚弄完拆绝对值的，今天弄拆最大值的。

默认法杖为 $0$ ，咒语为 $1$ 。如果 $a_0+a_1>b_0+b_1$ ，那么有 $a_0-b_0>b_1-a_1$ 。所以，对每个物品维护一个价值 $w$ 。

{\begin{matrix} w = a - b & c = 0 \\ w = b - a & c = 1 \end{matrix}

$\left\{\begin{matrix} w=a-b & c=0\\ w=b-a & c=1 \end{matrix}\right.$

那么就可以把这个 $w$ 放到权值线段树上去，同时对法杖和咒语分别维护一下区间最小的 $a$ 和 $b$ 。再稍微 pushup 一下即可。对于删除操作，使用 mutiset 维护线段树节点即可。注：卡空。

 #include<bits/stdc++.h>
using namespace std;
const int N = 5e5 + 100, inf = 1e8, M = 25e4;
int Q, T, ans, rt;
namespace ST{
	int cnt, ls[N<<1], rs[N<<1], l[N<<1], r[N<<1], num[N<<1], A[N<<1][2], B[N<<1][2];
	multiset<int> ta[N][2], tb[N][2];
	inline void pushup(int id){
		A[id][0] = min(A[ls[id]][0], A[rs[id]][0]), A[id][1] = min(A[ls[id]][1], A[rs[id]][1]);
		B[id][0] = min(B[ls[id]][0], B[rs[id]][0]), B[id][1] = min(B[ls[id]][1], B[rs[id]][1]);
		num[id] = min({num[ls[id]], num[rs[id]], A[ls[id]][1] + A[rs[id]][0], B[ls[id]][0] + B[rs[id]][1]});
	}
	inline void build(int &id, int x, int y){
		if(!id) id = ++cnt;
		l[id] = x, r[id] = y, num[id] = inf;
		A[id][0] = A[id][1] = B[id][0] = B[id][1] = inf;
		if(x == y){
			ta[x][0].insert(inf), ta[x][1].insert(inf);
			tb[x][0].insert(inf), tb[x][1].insert(inf);
			return;
		} int mid = (x + y) >> 1;
		build(ls[id], x, mid), build(rs[id], mid+1, y);
	}
	inline void add(int id, int ps, bool k, int a, int b){
		if(l[id] == r[id]){
			ta[ps][k].insert(a), tb[ps][k].insert(b);
			num[id] = min((*ta[ps][0].begin()) + (*ta[ps][1].begin()), (*tb[ps][0].begin()) + (*tb[ps][1].begin()));
			A[id][k] = *ta[ps][k].begin(); B[id][k] = *tb[ps][k].begin();
			return;
		} int mid = (l[id] + r[id]) >> 1;
		add(((ps <= mid) ? ls[id] : rs[id]), ps, k, a, b);
		pushup(id);
	}
	inline void del(int id, int ps, bool k, int a, int b){
		if(l[id] == r[id]){
			if(ta[ps][k].find(a) == ta[ps][k].end()) return;
			ta[ps][k].erase(ta[ps][k].find(a)), tb[ps][k].erase(tb[ps][k].find(b));
			num[id] = min((*ta[ps][0].begin()) + (*ta[ps][1].begin()), (*tb[ps][0].begin()) + (*tb[ps][1].begin()));
			A[id][k] = *ta[ps][k].begin(); B[id][k] = *tb[ps][k].begin();
			return;
		} int mid = (l[id] + r[id]) >> 1;
		del(((ps <= mid) ? ls[id] : rs[id]), ps, k, a, b);
		pushup(id);
	}
}
int main(){
//	freopen("30.in", "r", stdin), freopen("out.out", "w", stdout);
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin>>Q>>T; ST::build(rt, 1, 5e5+1);
	for(int i=1, opt, t, a, b; i<=Q; ++i){
		cin>>opt>>t>>a>>b;
		if(T) a ^= ans, b ^= ans;
		if(opt != 2) ST::add(rt, (t ? b-a+M : a-b+M), t, a, b);
		else ST::del(rt, (t ? b-a+M : a-b+M), t, a, b);
		ans = ST::num[rt] >= inf ? 0 : ST::num[rt];
		cout<<ans<<'\n';
	} return 0;
}

T4 园艺

机房大佬用拐两次的方法骗分直接AC……

30pts

很典的区间DP，设 $dp[l][r]$ 和 $dp[r][l]$ 分别表示拔完 $l\sim r$ 的草的价值并停留在 $l$ 和 $r$ 。 $\mathcal{O}(n^2)$ 。

100pts

上面的dp无法继续优化，考虑换一种计算贡献的形式。

考虑这么一个事，如果从 $k$ 点直接前往第 $p$ 株草可以获得 $dis(k,p)$ 的贡献，对于每一株草来说，这个贡献是最基础的，不能再少了。如果刚开始时从 $k$ 向左走 $m$ 步，那么在 $k$ 右侧的草的贡献

 #include<bits/stdc++.h>
using namespace std;
constexpr int B = 1 << 23;
char buf[B], *p1 = buf, *p2 = buf, obuf[B], *O = obuf;
#define gt() (p1==p2 && (p2=(p1=buf) + fread(buf, 1, B, stdin), p1==p2) ? EOF : *p1++)
template <typename T> inline void rd(T &x){
	x = 0; int f = 0; char ch = gt();
	for(; !isdigit(ch); ch = gt()) f ^= ch == '-';
	for(; isdigit(ch); ch = gt()) x = (x<<1) + (x<<3) + (ch^48);
	x = f ? -x : x;
}
#define pt(ch) (O-obuf==B && (fwrite(obuf, 1, B, stdout), O=obuf), *O++ = (ch))
template <typename T> inline void wt(T x){
	if(x < 0) pt('-'), x = -x;
	if(x >= 10) wt(x / 10); pt(x % 10 ^ 48);
}
#define fw fwrite(obuf, 1, O - obuf, stdout)
#define int long long
#define it __int128
#define p pair<int, int>
constexpr int N = 2e6 + 5;
int n, k, s[N], dp[N]; 
struct XiaoLe{
	int tail, head; p st[N];
	inline bool check(p a, p b, p c){
		return ((it)(b.second - a.second) * (c.first - b.first) > (it)(c.second - b.second) * (b.first - a.first)) ? 1 : 0;
	}
	inline void push_back(p node){
		while(head < tail && check(st[tail-1], st[tail], node)) --tail;
		st[++tail] = node;
	}
	inline void push_front(p node){
		while(head < tail && check(node, st[head], st[head+1])) ++head;
		st[--head] = node;
	}
	inline p get_ltor(int k){
		while(head < tail && (it)(st[tail].second - st[tail-1].second) > (it)k * (st[tail].first - st[tail-1].first)) --tail;
		return st[tail];
	}
	inline p get_rtol(int k){
		while(head < tail && (it)(st[head+1].second - st[head].second) < (it)k * (st[head+1].first - st[head].first)) ++head;
		return st[head]; 
	}
}xl[2]; // 0 : l->r 1 : r->l
signed main(){
	rd(n), rd(k);
	for(int i=2, a; i<=n; ++i) rd(a), s[i] = s[i-1] + a;
	memset(dp, 0x7f, sizeof(int) * (n+1)); dp[k] = 0;
	for(int i=1; i<=n; ++i) dp[k] += llabs(s[k] - s[i]);
	int tot = n - 1, l = k, r = k;
	xl[0].head = xl[1].head = k; xl[0].tail = xl[1].tail = k - 1;
	xl[0].push_back(p{k, dp[k] + 2*(n-k)*s[k]});
	xl[1].push_front(p{k, dp[k] - 2*(k-1)*s[k]});
	while(tot--){
		if(l == 1 || r == n) break;
		p ans = xl[0].get_rtol(-2*s[l-1]);
		dp[l-1] = ans.second + 2*ans.first*s[l-1] - 2*n*s[l-1];
		ans = xl[1].get_ltor(-2*s[r+1]);
		dp[r+1] = ans.second + 2*ans.first*s[r+1] - 2*s[r+1];
		if(dp[l-1] <= dp[r+1]) --l, xl[1].push_front({l, dp[l] - 2*(l-1)*s[l]});
		else ++r, xl[0].push_back({r, dp[r] + 2*(n-r)*s[r]});
	} wt(min(dp[1], dp[n])); return fw, 0;
}

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本文作者：XiaoLe_MC

本文链接：https://www.cnblogs.com/xiaolemc/p/18425910

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[考试记录] 2024.9.16 csp-s模拟赛30

T1 不相邻集合

T2 线段树

solution1

solution 2

solution 3

正解

T3 魔法师

T4 园艺

30pts

100pts

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	#include<bits/stdc++.h>
	using namespace std;
	constexpr int B = 1 << 25;
	char buf[B], p1 = buf, p2 = buf;
	#define gt() (p1==p2 && (p2=(p1=buf)+fread(buf, 1, B, stdin), p1==p2) ? EOF : *p1++)
	template <typename T> inline void rd(T &x){
	x = 0; int f = 0; char ch = gt();
	for(; !isdigit(ch); ch = gt()) f ^= ch == '-';
	for(; isdigit(ch); ch = gt()) x = (x<<1) + (x<<3) + (ch^48);
	x = f ? -x : x;
	}
	char obuf[B], *O = obuf;
	#define pt(ch) (O-obuf==B && (fwrite(obuf, 1, B, stdout), O=obuf), *O++ = (ch))
	template <typename T> inline void wt(T x){
	if(x < 0) pt('-'), x = -x;
	if(x >= 10) wt(x / 10); pt(x % 10 ^ 48);
	}
	#define fw fwrite(obuf, 1, O - obuf, stdout)
	constexpr int N = 3e5 + 5, T = 1e6 + 5, M = 5e5 + 1;
	int n, x, ans, mx, rt, as1, as2;
	namespace ST{
	int cnt, ls[T], rs[T], f[T], g[T];
	inline void fadd(int &id, int l, int r, int x, int val){
	if(!id) id = ++cnt;
	if(l == r){
	f[id] = max(f[id], val);
	return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) fadd(ls[id], l, mid, x, val);
	else fadd(rs[id], mid+1, r, x, val);
	f[id] = max(f[ls[id]], f[rs[id]]);
	}
	inline void gadd(int &id, int l, int r, int x, int val){
	if(!id) id = ++cnt;
	if(l == r){
	g[id] = max(g[id], val);
	return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) gadd(ls[id], l, mid, x, val);
	else gadd(rs[id], mid+1, r, x, val);
	g[id] = max(g[ls[id]], g[rs[id]]);
	}
	inline int fquery(int &id, int l, int r, int x, int y){
	if(!id \|\| x > y) return 0;
	if(x <= l && r <= y) return f[id];
	int mid = (l + r) >> 1, ans = 0;
	if(x <= mid) ans = max(ans, fquery(ls[id], l, mid, x, y));
	if(y > mid) ans = max(ans, fquery(rs[id], mid+1, r, x, y));
	return ans;
	}
	inline int gquery(int &id, int l, int r, int x, int y){
	if(!id \|\| x > y) return 0;
	if(x <= l && r <= y) return g[id];
	int mid = (l + r) >> 1, ans = 0;
	if(x <= mid) ans = max(ans, gquery(ls[id], l, mid, x, y));
	if(y > mid) ans = max(ans, gquery(rs[id], mid+1, r, x, y));
	return ans;
	}
	}
	bitset<M> vis;
	int main(){
	rd(n); while(n--){
	rd(x); if(vis[x]){ wt(ans), pt(' '); continue; }
	vis[x] = 1;
	as1 = ST::fquery(rt, 1, M, 1, x-2) + 1;
	ST::fadd(rt, 1, M, x, as1);
	as2 = ST::gquery(rt, 1, M, x+2, M) + 1;
	ST::gadd(rt, 1, M, x, as2);
	ans = max(ans, as1 + as2 - 1);
	wt(ans), pt(' ');
	} return fw, 0;
	}

	#include<bits/stdc++.h>
	using namespace std;
	const int N = 5e5 + 100, inf = 1e8, M = 25e4;
	int Q, T, ans, rt;
	namespace ST{
	int cnt, ls[N<<1], rs[N<<1], l[N<<1], r[N<<1], num[N<<1], A[N<<1][2], B[N<<1][2];
	multiset<int> ta[N][2], tb[N][2];
	inline void pushup(int id){
	A[id][0] = min(A[ls[id]][0], A[rs[id]][0]), A[id][1] = min(A[ls[id]][1], A[rs[id]][1]);
	B[id][0] = min(B[ls[id]][0], B[rs[id]][0]), B[id][1] = min(B[ls[id]][1], B[rs[id]][1]);
	num[id] = min({num[ls[id]], num[rs[id]], A[ls[id]][1] + A[rs[id]][0], B[ls[id]][0] + B[rs[id]][1]});
	}
	inline void build(int &id, int x, int y){
	if(!id) id = ++cnt;
	l[id] = x, r[id] = y, num[id] = inf;
	A[id][0] = A[id][1] = B[id][0] = B[id][1] = inf;
	if(x == y){
	ta[x][0].insert(inf), ta[x][1].insert(inf);
	tb[x][0].insert(inf), tb[x][1].insert(inf);
	return;
	} int mid = (x + y) >> 1;
	build(ls[id], x, mid), build(rs[id], mid+1, y);
	}
	inline void add(int id, int ps, bool k, int a, int b){
	if(l[id] == r[id]){
	ta[ps][k].insert(a), tb[ps][k].insert(b);
	num[id] = min((ta[ps][0].begin()) + (ta[ps][1].begin()), (tb[ps][0].begin()) + (tb[ps][1].begin()));
	A[id][k] = ta[ps][k].begin(); B[id][k] = tb[ps][k].begin();
	return;
	} int mid = (l[id] + r[id]) >> 1;
	add(((ps <= mid) ? ls[id] : rs[id]), ps, k, a, b);
	pushup(id);
	}
	inline void del(int id, int ps, bool k, int a, int b){
	if(l[id] == r[id]){
	if(ta[ps][k].find(a) == ta[ps][k].end()) return;
	ta[ps][k].erase(ta[ps][k].find(a)), tb[ps][k].erase(tb[ps][k].find(b));
	num[id] = min((ta[ps][0].begin()) + (ta[ps][1].begin()), (tb[ps][0].begin()) + (tb[ps][1].begin()));
	A[id][k] = ta[ps][k].begin(); B[id][k] = tb[ps][k].begin();
	return;
	} int mid = (l[id] + r[id]) >> 1;
	del(((ps <= mid) ? ls[id] : rs[id]), ps, k, a, b);
	pushup(id);
	}
	}
	int main(){
	// freopen("30.in", "r", stdin), freopen("out.out", "w", stdout);
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin>>Q>>T; ST::build(rt, 1, 5e5+1);
	for(int i=1, opt, t, a, b; i<=Q; ++i){
	cin>>opt>>t>>a>>b;
	if(T) a ^= ans, b ^= ans;
	if(opt != 2) ST::add(rt, (t ? b-a+M : a-b+M), t, a, b);
	else ST::del(rt, (t ? b-a+M : a-b+M), t, a, b);
	ans = ST::num[rt] >= inf ? 0 : ST::num[rt];
	cout<<ans<<'\n';
	} return 0;
	}

	#include<bits/stdc++.h>
	using namespace std;
	constexpr int B = 1 << 23;
	char buf[B], p1 = buf, p2 = buf, obuf[B], *O = obuf;
	#define gt() (p1==p2 && (p2=(p1=buf) + fread(buf, 1, B, stdin), p1==p2) ? EOF : *p1++)
	template <typename T> inline void rd(T &x){
	x = 0; int f = 0; char ch = gt();
	for(; !isdigit(ch); ch = gt()) f ^= ch == '-';
	for(; isdigit(ch); ch = gt()) x = (x<<1) + (x<<3) + (ch^48);
	x = f ? -x : x;
	}
	#define pt(ch) (O-obuf==B && (fwrite(obuf, 1, B, stdout), O=obuf), *O++ = (ch))
	template <typename T> inline void wt(T x){
	if(x < 0) pt('-'), x = -x;
	if(x >= 10) wt(x / 10); pt(x % 10 ^ 48);
	}
	#define fw fwrite(obuf, 1, O - obuf, stdout)
	#define int long long
	#define it __int128
	#define p pair<int, int>
	constexpr int N = 2e6 + 5;
	int n, k, s[N], dp[N];
	struct XiaoLe{
	int tail, head; p st[N];
	inline bool check(p a, p b, p c){
	return ((it)(b.second - a.second) * (c.first - b.first) > (it)(c.second - b.second) * (b.first - a.first)) ? 1 : 0;
	}
	inline void push_back(p node){
	while(head < tail && check(st[tail-1], st[tail], node)) --tail;
	st[++tail] = node;
	}
	inline void push_front(p node){
	while(head < tail && check(node, st[head], st[head+1])) ++head;
	st[--head] = node;
	}
	inline p get_ltor(int k){
	while(head < tail && (it)(st[tail].second - st[tail-1].second) > (it)k * (st[tail].first - st[tail-1].first)) --tail;
	return st[tail];
	}
	inline p get_rtol(int k){
	while(head < tail && (it)(st[head+1].second - st[head].second) < (it)k * (st[head+1].first - st[head].first)) ++head;
	return st[head];
	}
	}xl[2]; // 0 : l->r 1 : r->l
	signed main(){
	rd(n), rd(k);
	for(int i=2, a; i<=n; ++i) rd(a), s[i] = s[i-1] + a;
	memset(dp, 0x7f, sizeof(int) * (n+1)); dp[k] = 0;
	for(int i=1; i<=n; ++i) dp[k] += llabs(s[k] - s[i]);
	int tot = n - 1, l = k, r = k;
	xl[0].head = xl[1].head = k; xl[0].tail = xl[1].tail = k - 1;
	xl[0].push_back(p{k, dp[k] + 2(n-k)s[k]});
	xl[1].push_front(p{k, dp[k] - 2(k-1)s[k]});
	while(tot--){
	if(l == 1 \|\| r == n) break;
	p ans = xl[0].get_rtol(-2*s[l-1]);
	dp[l-1] = ans.second + 2ans.firsts[l-1] - 2ns[l-1];
	ans = xl[1].get_ltor(-2*s[r+1]);
	dp[r+1] = ans.second + 2ans.firsts[r+1] - 2*s[r+1];
	if(dp[l-1] <= dp[r+1]) --l, xl[1].push_front({l, dp[l] - 2(l-1)s[l]});
	else ++r, xl[0].push_back({r, dp[r] + 2(n-r)s[r]});
	} wt(min(dp[1], dp[n])); return fw, 0;
	}