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建出来笛卡尔树
一个子树是原序列上的一个区间，所以可以预处理出 ST 表
然后二分一下两个子树中最小值 \(>x-a_u\) 和 \(\geqslant x-a_u\) 的位置就好了
复杂度 \(O(n\log n)\)，听说也有 \(O(n)\) 做法

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#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define fir first
#define sec second
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, x;
int a[N];

// namespace force{
// 	ll ans;
// 	int st1[22][N], st2[22][N], lg[N];
// 	inline int qmin(int l, int r) {
// 		int t=lg[r-l+1]-1;
// 		return min(st1[t][l], st1[t][r-(1<<t)+1]);
// 	}
// 	inline int qmax(int l, int r) {
// 		int t=lg[r-l+1]-1;
// 		return max(st2[t][l], st2[t][r-(1<<t)+1]);
// 	}
// 	void solve() {
// 		for (int i=1; i<=n; ++i) lg[i]=lg[i-1]+(1<<lg[i-1]==i);
// 		for (int i=1; i<=n; ++i) st1[0][i]=a[i];
// 		int t=lg[n]-1;
// 		for (int i=1; i<=t; ++i)
// 			for (int j=1; j+(1<<i)-1<=n; ++j)
// 				st1[i][j]=min(st1[i-1][j], st1[i-1][j+(1<<i-1)]);
// 		for (int i=1; i<=n; ++i) st2[0][i]=a[i];
// 		for (int i=1; i<=t; ++i)
// 			for (int j=1; j+(1<<i)-1<=n; ++j)
// 				st2[i][j]=max(st2[i-1][j], st2[i-1][j+(1<<i-1)]);
// 		for (int l=1; l<=n; ++l)
// 			for (int r=l; r<=n; ++r)
// 				if (qmax(l, r)+qmin(l, r)==x)
// 					++ans;
// 		printf("%lld\n", ans);
// 	}
// }

namespace task1{
	ll ans;
	pair<int, int> sta[N];
	int ls[N], rs[N], st[22][N], lg[N], siz[N], top;
	inline int qmin(int l, int r) {
		int t=lg[r-l+1]-1;
		return min(st[t][l], st[t][r-(1<<t)+1]);
	}
	void dfs(int u) {
		siz[u]=1;
		if (ls[u]) dfs(ls[u]), siz[u]+=siz[ls[u]];
		if (rs[u]) dfs(rs[u]), siz[u]+=siz[rs[u]];
		int l, r, mid;
		int l_l, l_r, r_l, r_r;

		// cout<<"calc: "<<u<<endl;

		l=u+1, r=u+siz[rs[u]];
		while (l<=r) {
			mid=(l+r)>>1;
			if (qmin(u, mid)>x-a[u]) l=mid+1;
			else r=mid-1;
		}
		r_l=l-1;

		l=u+1, r=u+siz[rs[u]];
		while (l<=r) {
			mid=(l+r)>>1;
			if (qmin(u, mid)>=x-a[u]) l=mid+1;
			else r=mid-1;
		}
		r_r=l-1;

		l=u-siz[ls[u]], r=u-1;
		while (l<=r) {
			mid=(l+r)>>1;
			if (qmin(mid, u)>=x-a[u]) r=mid-1;
			else l=mid+1;
		}
		l_l=r+1;

		l=u-siz[ls[u]], r=u-1;
		while (l<=r) {
			mid=(l+r)>>1;
			if (qmin(mid, u)>x-a[u]) r=mid-1;
			else l=mid+1;
		}
		l_r=r+1;

		// cout<<"range: "<<l_l<<' '<<l_r<<' '<<r_l<<' '<<r_r<<endl;

		if (x-a[u]<=0) {/* cout<<"case1"<<endl; */ return ;}
		else if (x-a[u]==a[u]) ans+=1ll*(u-l_l+1)*(r_r-u+1); //, cout<<"case2"<<endl;
		else ans+=1ll*(u-l_l+1)*(r_r-u+1)-1ll*(u-l_r+1)*(r_l-u+1); //, cout<<"case3"<<endl;
	}
	void solve() {
		for (int i=1; i<=n; ++i) lg[i]=lg[i-1]+(1<<lg[i-1]==i);
		for (int i=1; i<=n; ++i) st[0][i]=a[i];
		int t=lg[n]-1;
		for (int i=1; i<=t; ++i)
			for (int j=1; j+(1<<i)-1<=n; ++j)
				st[i][j]=min(st[i-1][j], st[i-1][j+(1<<i-1)]);
		for (int i=1; i<=n; ++i) {
			pair<int, int> t={a[i], i};
			int k=top;
			while (k && sta[k].fir<t.fir) --k;
			if (k!=top) ls[t.sec]=sta[k+1].sec;
			if (k) rs[sta[k].sec]=t.sec;
			sta[top=++k]=t;
		}
		// cout<<"ls: "; for (int i=1; i<=n; ++i) cout<<ls[i]<<' '; cout<<endl;
		// cout<<"rs: "; for (int i=1; i<=n; ++i) cout<<rs[i]<<' '; cout<<endl;
		dfs(sta[1].sec);
		printf("%lld\n", ans);
	}
}

signed main()
{
	freopen("garlic.in", "r", stdin);
	freopen("garlic.out", "w", stdout);

	n=read(); x=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	// force::solve();
	task1::solve();

	return 0;
}