[树状数组] ARC075 E - Meaningful Mean

$Solution$

设$a_i$的前缀和为$S_i$。满足条件的区间$[l,r]$相当于${S_r-S_l\over r-l}\ge k$，整理一下就是$s_r-kr\ge s_l-kl$。
设$f_i=s_i-ki$，题目求得就是$f_l\le f_r,l\lt r$的点对$(l,r)$个数。
树状数组。
哦。。其实单调栈就行了吧

#include <bits/stdc++.h>
using namespace std;

const int N = 202020;
typedef long long ll;

inline char get(void) {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
template<typename T>
inline void read(T &x) {
  static char c; x = 0; int sgn = 0;
  for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
  if (sgn) x = -x;
}

int a[N], g[N], c[N];
ll s[N], f[N], mp[N];
ll ans;
int n, m, k;

inline void Add(int x) {
  for (; x <= m; x += x & -x) ++c[x];
}
inline int Sum(int x) {
  int sum = 0;
  for (; x; x -= x & -x) sum += c[x];
  return sum;
}

int main(void) {
  freopen("1.in", "r", stdin);
  freopen("1.out", "w", stdout);
  read(n); read(k);
  for (int i = 1; i <= n; i++) {
    read(a[i]); s[i] = s[i - 1] + a[i];
  }
  for (int i = 0; i <= n; i++)
    mp[i] = f[i] = s[i] - (ll)k * i;
  sort(mp, mp + n + 1);
  m = unique(mp, mp + n + 1) - mp;
  for (int i = 0; i <= n; i++)
    g[i] = lower_bound(mp, mp + m, f[i]) - mp + 1;
  for (int i = 0; i <= n; i++) {
    ans += Sum(g[i]); Add(g[i]);
  }
  cout << ans << endl;
  return 0;
}