codeforces B. XK Segments

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最新推荐文章于 2025-01-02 00:05:29 发布

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解决一个算法问题，找出数组中满足特定条件的有序数对数量。对于数组a和整数x，找到所有满足ai≤aj且恰好有k个整数y（ai≤y≤aj且y能被x整除）的不同有序数对(i,j)。

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While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array a and integer x. He should find the number of different ordered pairs of indexes (i, j) such that ai ≤ aj and there are exactly k integers y such that ai ≤ y ≤ aj and y is divisible by x.

In this problem it is meant that pair (i, j) is equal to (j, i) only if i is equal to j. For example pair (1, 2) is not the same as (2, 1).

Input

The first line contains 3 integers n, x, k (1 ≤ n ≤ 105, 1 ≤ x ≤ 109, 0 ≤ k ≤ 109), where n is the size of the array a and x and k are numbers from the statement.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

Output

Print one integer — the answer to the problem.

Examples

input
4 2 1
1 3 5 7

output
3

input
4 2 0
5 3 1 7

output
4

input
5 3 1
3 3 3 3 3

output
25

Note

In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4).

In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4).

In third sample every pair (i, j) is suitable, so the answer is 5 * 5 = 25.

思路：先对整体从小到大排序，然后枚举每个a[i],寻找符合条件的上限值和下限值，然后用lower_bound求这个a[i]的方案数

#include<iostream>
#include<cstdlib>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define maxn 100005
#define INF 1000000009
using namespace std;
long long n,x,k;
long long a[maxn];
int main()
{
	ios_base::sync_with_stdio(false);
	while(cin>>n)
	{
		cin>>x>>k;
		for(int i=0;i<n;i++)
		{
			cin>>a[i]; 
		}
		sort(a,a+n);
		long long cnt=0;
		for(int i=0;i<n;i++)
		{
			long long l=max(((a[i]-1)/x+k)*x,a[i]);//下限,当k=0时，a[i]可能更大，所以用max 
			long long r=((a[i]-1)/x+k+1)*x;//上限 
			cnt+=lower_bound(a,a+n,r)-lower_bound(a,a+n,l);
		}
		cout << cnt << endl;
	}
	return 0;
}