写了个开区间二分，感觉在需要函数二分时比较好理解

#include <bits/stdc++.h>
using namespace std;
#define int long long
int t, n, k, sum;

void solve() {
    cin >> n >> k;
    int l = max(-1ll, n / 2 - 1), r = n;
    const int sum = (2 * k + n - 1) * n / 2;
    auto f = [&](int m) {
        return (2 * k + m) * (m + 1) - sum;
    };
    while (r - l > 1) {
        int mid = l + (r - l) / 2;
        if (f(mid) < 0) {
            l = mid;
        } else if (f(mid) > 0) {
            r = mid;
        } else {
            break;
        }
    }
    l = max(0ll, l), r = min(r, n - 1);
    cout << min(abs(f(l)), abs(f(r))) << '\n';
}

signed main() {
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}