思路：

此题是求有多少个区间的平均值>=t， 那么可以把每个值-t。如果新的数列的某个区间的和>=0，那么说明这个区间满足条件。

令新数列的前缀和为b[i]，所以求[i, j]区间是否满足条件，即求b[j]-b[i-1]是否>=0，即b[j]>=b[i-1]。

因为j>i>i-1，所以这里即求“伪逆序对”的数量。

扩展知识：

逆序对：i>j a[i]<a[j]      伪逆序对/非逆序对：i>j a[i]>a[j]

方法：归并排序

代码：

1.8/10代码：错误原因：超时

#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e6 + 10;
long long int p = 1e9 + 7;
long long int n, t;
long long int a[N];
long long int b[N];
int main()
{cin >> n >> t;for (long long int i = 1; i <= n; i++){cin >> a[i];a[i] -= t;}for (long long int i = 1; i <= n; i++){b[i] = b[i - 1] + a[i];}long long int ans = 0;for (long long int i = 1; i <= n; i++){for (long long int j = 1; j <= i; j++){if (b[i] - b[j - 1] >= 0){ans++;}}}cout << ans % p;
}

2.10/10代码：升序排列求逆序对，再用总的-逆序对即为非逆序对个数

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e6 + 10;
int p = 1e9 + 7;
ll n, t;
ll a[N], sum[N], q[N];
ll ans = 0;
void merge_sort(int l, int r, ll a[])
{if (l >= r)return;int mid = (l + r) >> 1;merge_sort(l, mid, a);merge_sort(mid + 1, r, a);int i = l, j = mid + 1, k = 0;while (i <= mid && j <= r){if (a[i] > a[j]){q[k++] = a[j++];ans += mid - i + 1; // 升序排列，求逆序数ans %= p;}else{q[k++] = a[i++];}}while (i <= mid)q[k++] = a[i++];while (j <= r)q[k++] = a[j++];for (i = l, j = 0; i <= r; i++, j++){a[i] = q[j];}
}int main()
{cin >> n >> t;for (int i = 1; i <= n; i++){cin >> a[i];a[i] -= t;sum[i] = sum[i - 1] + a[i];}merge_sort(0, n, sum);cout << (n * (n + 1) / 2 - ans) % p;return 0;
}

3.10/10代码，直接降序求非逆序对个数

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e6 + 10;
int p = 1e9 + 7;
ll n, t;
ll a[N], sum[N], q[N];
ll ans = 0;
void merge_sort(int l, int r, ll a[])
{if (l >= r)return;int mid = (l + r) >> 1;merge_sort(l, mid, a);merge_sort(mid + 1, r, a);int i = l, j = mid + 1, k = 0;while (i <= mid && j <= r){if (a[i] <= a[j]){q[k++] = a[j++];ans += mid - i + 1; // 降序排列，求非逆序数ans %= p;}else{q[k++] = a[i++];}}while (i <= mid)q[k++] = a[i++];while (j <= r)q[k++] = a[j++];for (i = l, j = 0; i <= r; i++, j++){a[i] = q[j];}
}int main()
{cin >> n >> t;for (int i = 1; i <= n; i++){cin >> a[i];a[i] -= t;sum[i] = sum[i - 1] + a[i];}merge_sort(0, n, sum);cout << ans % p;return 0;
}