BZOJ 1303: [CQOI2009]中位数图

要求左右选出来的数满足，比 \(b\) 大的等于比 \(b\) 小的
从 \(b\) 向左右分别拓展
向左遇到一个比 \(b\) 大的就 -1，比 \(b\) 小的就 +1
向右拓展就反过来统计
最后答案就是 \(\sum left_i \times right_i\)

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define SZ(x) ((int)(x).size())
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=((b)-1);i>=(a);i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1000000007, INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2, int mod = MOD) {int ans = 1; for (; b; a = 1LL * a * a % mod, b >>= 1)if (b & 1)ans = 1LL * ans * a % mod; return ans % mod;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 1e5 + 7;
int le[2 * N], ri[2 * N], a[N], b, n;

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	n = _(), b = _();
	int pos = 0;
	rep (i, 1, n + 1) {
		a[i] = _();
		if (a[i] == b) pos = i;
	}
	int cur = 0;
	per (i, 0, pos) {
		le[cur + N]++;
		if (a[i] < b) cur--;
		else cur++;
	}
	cur = 0;
	rep (i, pos + 1, n + 2) {
		ri[cur + N]++;
		if (a[i] < b) cur++;
		else cur--;
	}
	ll ans = 0;
	rep (i, 0, 2 * N) ans += (ll)le[i] * ri[i];
	printf("%lld\n", ans);
#ifdef LOCAL
	printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
	return 0;
}