Codeforces Round #448 (Div. 2)B

B. XK Segments
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
자기가 진짜 못하네.
이 문제는 오랫동안 생각해서 다른 사람의 코드를 보고 나서야 알아차렸다
만족하는 구간을 찾아 각각 그와 인접한 하표를 찾아 상쇄하다

#include
using namespace std;
#define maxn 1000000+100
#define ll long long
long long a[maxn];
int main(){
   ll n,k,x;
   scanf("%lld%lld%lld",&n,&x,&k);
   for(ll j=0;j

#include
using namespace std;
#define maxn 1000000+100
#define ll long long
long long a[maxn];
ll binary_search1(ll l,ll r,ll x){
    while(l<=r){
        int mid=(l+r)/2;
        if(a[mid]>=x){
             r=mid-1;
        }
        else l=mid+1;
    }
    return l;
}
int main(){
   ll n,k,x;
   scanf("%lld%lld%lld",&n,&x,&k);
   for(ll j=0;j