Algorithm


Problem link- https://www.spoj.com/problems/KQUERYO/

KQUERYO - K-Query Online

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Given a sequence of n numbers a1, a2, ..., an and a number of k-queries. A k-query is a triple (i, j, k) (1 ≤ i ≤ j ≤ n). For each k-query (i, j, k), you have to return the number of elements greater than k in the subsequence ai, ai+1, ..., aj.

Input

  • Line 1: n (1 ≤ n ≤ 30000).
  • Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
  • Line 3: q (1 ≤ q ≤ 200000), the number of k- queries.
  • In the next q lines, each line contains 3 numbers a, b, c representing a k-query. You should do the following:
    • i = a xor last_ans
    • j = b xor last_ans
    • k = c xor last_ans
    After that 1 ≤ i ≤ j ≤ n, 1 ≤ k ≤ 109 holds.
    Where last_ans = the answer to the last query (for the first query it's 0).

Output

For each k-query (i, j, k), print the number of elements greater than k in the subsequence ai, ai+1, ..., aj in a single line.

Example

Input:
6
8 9 3 5 1 9
5
2 3 5
3 3 7
0 0 11
0 0 2
3 7 4

Output:
1
1
0
0
2

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
#define MOD                 1000000007LL
#define EPS                 1e-9
#define io                  ios_base::sync_with_stdio(false);cin.tie(NULL);
#define M_PI                3.14159265358979323846

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

const int MAXN = 1e5+5;
int a[MAXN];
vector<int> b[320];

int main(){
    io;
    int n;
    cin >> n;
    int sqrtn = sqrt(n);
    for(int i = 0;i < n; i++){
    	cin >> a[i];
    	b[i/sqrtn].push_back(a[i]);
    }
    int numBlocks = ceil((double)n/sqrtn);
    for(int i = 0;i < numBlocks; i++)
    	sort(b[i].begin(), b[i].end());
    int q;
    cin >> q;
    int answer = 0;
    while(q--){
    	int l, r, x;
    	cin >> l >> r >> x;
    	l ^= answer;
    	r ^= answer;
    	x ^= answer;
    	if(l < 1)
    		l = 1;
    	if(r > n)
    		r = n;
    	l--;r--;
	    answer = 0;
    	if(r >= l){
	    	int leftBlock = l/sqrtn;
	    	int rightBlock = r/sqrtn;
	    	if(leftBlock == rightBlock){
	    		for(int i = l; i <= r; i++){
	    			if(a[i] > x)
	    				answer++;
	    		}
	    	}else{
	    		if(l%sqrtn != 0){
	    			leftBlock++;
	    		}
	    		int i;
	    		for(i = l; i < leftBlock*sqrtn; i++){
	    			if(a[i] > x)
	    				answer++;
	    		}
	    		while(i+sqrtn-1 <= r){
	    			int bb = i/sqrtn;
	    			answer += b[bb].end() - upper_bound(b[bb].begin(), b[bb].end(), x);
	    			i += sqrtn;
	    		}
	    		while(i <= r){
	    			if(a[i] > x)
	    				answer++;
	    			i++;
	    		}
	    	}
	    }
	    cout << answer << endl;
    }
    return 0;
}
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Input

x
+
cmd
6
8 9 3 5 1 9
5
2 3 5
3 3 7
0 0 11
0 0 2
3 7 4

Output

x
+
cmd
1
1
0
0
2
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Demonstration


SPOJ Solution-K-Query Online-Solution in C, C++, Java, Python

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