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最新推荐文章于 2022-05-31 22:37:57 发布

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最新推荐文章于 2022-05-31 22:37:57 发布

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原文链接：http://www.cnblogs.com/--lr/p/7922647.html

本文介绍了一种算法问题，即在一个整数数组中找到满足特定条件的有序数对(i, j)，并给出了详细的输入输出样例及解决方案。该算法通过排序和二分查找实现了高效的求解。

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 a_i ≤ a_j and there are exactly k integers y such that a_i ≤ y ≤ a_j 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 a_i (1 ≤ a_i ≤ 10⁹) — the elements of the array a.

Output

Print one integer — the answer to the problem.

Example

Input

4 2 1
1 3 5 7

Output

Input

4 2 0
5 3 1 7

Output

Input

5 3 1
3 3 3 3 3

Output

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.

#include <cstdio>
#include <iostream>
#include <string.h>
#include <cmath>
#include <algorithm>

#define LL long long
using namespace std;

LL a[100000+10];
int main()
{
    int n,x,k;
    LL result = 0;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%d%d",&x,&k);
        result = 0;
        for(int i=1;i<=n;i++)cin>>a[i];
        sort(a+1,a+n+1);
        for(int i=1;i<=n;i++)
        {
            LL l = max(a[i],(k+((a[i]-1)/x))*x);
            LL r = max(a[i],(k+1+((a[i]-1)/x))*x);
            //cout<<l;
            //printf("    ");
            //cout<<r<<endl;
            result += (lower_bound(a+1,a+n+1,r)-lower_bound(a+1,a+n+1,l));
        }
        printf("%lld\n",result);
    }

    return 0;
}
//2   6 13   6-(3-1)
//6 8 10 12
//(k+((a[i]-1)/x))*x=a[j];
//a[j]/x-(a[i]-1)/x == k+1?