# POJ: K-th Number

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.

That is, given an array a[1…n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: “What would be the k-th number in a[i…j] segment, if this segment was sorted?”

For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2…5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n — — the size of the array, and m — — the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).

The second line contains n different integer numbers not exceeding 109 by their absolute values — — the array for which the answers should be given.

The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j — i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it — — the k-th number in sorted a[i…j] segment.

Sample Input

`7 3`

1 5 2 6 3 7 4

2 5 3

4 4 1

1 7 3

Sample Output

`5`

6

3

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

**Problem Analysis:**

Because m may be a large number, so if we sort the given segment then count the kth number, the time complexity would be O(mnlogn)

Now we optimize it. We can divide the array into some subarrays, support we divide the array into p subarrays, every subarray size is q, and n = pq. We can sort every subarrays. So if the given segment is [s, t], if it contains some subarrays, we can use binary search in these subarray, and at most 2 subarrays are not completed contained, so we count these arrays elements linearly.

**Solution**

#include <cstdio>

#include <vector>

#include <algorithm>using namespace std;#define B 707int n, m;int main(int argc, char* argv[]) {

scanf("%d%d", &n, &m);

vector<int> data(n);

vector<int> sorted(n);

int bucket_num = (n + B - 1) / B;

vector<vector<int> > buckets(bucket_num, vector<int>(B));

if (n % B != 0) buckets[bucket_num - 1].resize(n % B);

getchar();

for (int i = 0; i < n; ++i) {

scanf("%d", &data[i]);

sorted[i] = data[i];

buckets[i / B][i % B] = data[i];

}

sort(sorted.begin(), sorted.end());

for (int i = 0; i < bucket_num; ++i) sort(buckets[i].begin(), buckets[i].end()); while (m--) {

int i, j, k;

getchar();

scanf("%d%d%d", &i, &j, &k);

int start = -1;

int end = n - 1;

while (end > start + 1) {

int middle = start + (end - start) / 2;

int value = sorted[middle];

int c = 0;

int left = i - 1;

int right = j;

while (left < right && (left % B != 0))

if (data[left++] <= value) ++c;

while (left < right && (right % B != 0))

if (data[--right] <= value) ++c;

while (left < right) {

c += upper_bound(buckets[left / B].begin(), buckets[left / B].end(), value) - buckets[left / B].begin();

left += B;

} if (c >= k) end = middle;

else start = middle;

} printf("%d\n", sorted[end]);

} return 0;

}

time complexity is O(nlogn)

space complexity is O(n)