= query(0, i, j, sort[mid], 0, N);
if (query < k){
lf = mid;...* @param r
* @return
*/
public int query(int k, int i, int j, int x, int l, int r)...(dat[k], x) + 1;
}else{
int ans = 0;
ans += query(2 * k + 1, i, j, x,...(l + r) / 2;
Pair lch = query(2 * k + 1, i, j, l, mid);
Pair rch = query(2 *...BIT或线段树,BIT更简洁易懂。