80pts的逆天分数，#9#10TLE

#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;

ll A[11451419];
ll H[11451419];
ll i = 0;

ll work_out(ll x, ll y) {
    ll t1 = lower_bound(A + 1, A + 1 + i, x) - A;
    ll t2 = lower_bound(A + 1, A + 1 + i, y) - A;
    x -= A[t1 - 1];
    y -= A[t2 - 1];
    if (t1 == t2) {
        return H[y] - H[x - 1];
    }
    if (t1 == t2 - 1) {
        return H[t1] - H[x - 1] + H[y];
    }
    ll sum = 0;
    sum += H[t1] - H[x - 1];
    sum += H[y];
    t1++;
    while (t1 < t2) {
        sum += H[t1];
        t1++;
    }
    return sum;
}


int main() {
    ll T;
    scanf("%lld", &T);
    ll x, y;
    while (A[i] <= 1000000000000) {
        i++;
        A[i] = i * (i + 1) / 2; 
    }
    for (ll k = 1; k <= 1500000; k++) {
        H[k] = H[k - 1] + k;
    }
    ll t;
    while (T--) {
        scanf("%lld", &x);
        scanf("%lld", &y);
        t = work_out(x, y);
        printf("%lld\n", t);
    }
    return 0;
}