• 相向双指针
• 时间复杂度：$O(n)$
• 而且很好写，强烈推荐！！

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

using i64 = long long;

i64 solve(vector<int>& a, int m) {
    int n = a.size();
    i64 ans = 0;

    for (int l = 0, r = n; l < n; l++) {
        // a[r] 是最小的数使得 a[l] + a[r] >= m
        r = max(r, l + 1);
        while (r - 1 > l && a[r - 1] + a[l] >= m) r--;
        ans += n - r;
    }
    return ans;
}

int main() {	
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    cout << solve(a, m) << '\n';

    return 0;
}

• 二分做法
• 时间复杂度：$O(nlog(n))$

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

using i64 = long long;

i64 solve(vector<int>& a, int m) {
    int n = a.size();
    i64 ans = 0;

    for (int l = 0; l < n; l++) {
        // a[r] 是最小的数使得 a[l] + a[r] >= m
        int r = lower_bound(a.begin() + l + 1, a.end(), m - a[l]) - a.begin();
        ans += n - r;
    }
    return ans;
}

int main() {	
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    cout << solve(a, m) << '\n';

    return 0;
}