[CQOI2009]中位数

这道题的题意很显然。首先区间一定包括p，其次区间一定满足p左边右边大于p的数的个数和小于p的相等，于是将p左边和右边大于p和小于p的个数统计下。如果开二维数组记录空间太大了，事实上只要记录它们的差值即可。最后将差值能互补的乘起来加到答案里即可。具体做法见以下代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define inf (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 1e5 + 5;

int n, a[maxn], k, p, l[maxn << 1], r[maxn << 1];
int *ll = &l[maxn], *rr = &r[maxn];
int L = 0, R = 0;

LL sum = 0;

int main() {
    n = read(), k = read();
    rep(i, 1, n) {
        a[i] = read();
        if (a[i] == k) p = i;
    }
    ll[0] = rr[0] = 1;
    
    int x = 0;
    repd(i, p-1, 1) {
        if (a[i] > a[p]) x++; else x--;
        ll[x]++;
        maxx(R, x);
        minn(L, x);
    }
    x = 0;
    rep(i, p+1, n) {
        if (a[i] < a[p]) x++; else x--;
        rr[x]++;
    }

    rep(i, L, R) sum += (LL)ll[i] * rr[i];
    
    printf("%lld", sum);

    return 0;
}