CF - B -Micro-World - 代码天地

You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.

You know that you have $n$ bacteria in the Petri dish and size of the $i$ -th bacteria is $a_{i}$ . Also you know intergalactic positive integer constant $K$ .

The $i$ -th bacteria can swallow the $j$ -th bacteria if and only if $a_{i} > a_{j}$ and $a_{i} \leq a_{j} + K$ . The $j$ -th bacteria disappear, but the $i$ -th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_{i} > a_{j}$ and $a_{i} \leq a_{j} + K$ . The swallow operations go one after another.

For example, the sequence of bacteria sizes $a = [101, 53, 42, 102, 101, 55, 54]$ and $K = 1$ . The one of possible sequences of swallows is: $[101, 53, 42, 102, 101 - -- -, 55, 54]$ $\to$ $[101, 53 - - -, 42, 102, 55, 54]$ $\to$ $[101 - -- -, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, 54 - - -]$ $\to$ $[42, 102, 55]$ . In total there are $3$ bacteria remained in the Petri dish.

Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.

Input

The first line contains two space separated positive integers $n$ and $K$ ( $1 \leq n \leq 2 \cdot 10^{5}$ , $1 \leq K \leq 10^{6}$ ) — number of bacteria and intergalactic constant $K$ .

The second line contains $n$ space separated integers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 10^{6}$ ) — sizes of bacteria you have.

Output

Print the only integer — minimal possible number of bacteria can remain.

    Examples 
  
      input 
     
       Copy 
     
7 1
101 53 42 102 101 55 54

      output 
     
       Copy 
     
3

      input 
     
       Copy 
     
6 5
20 15 10 15 20 25

      output 
     
       Copy 
     
1

      input 
     
       Copy 
     
7 1000000
1 1 1 1 1 1 1

      output 
     
       Copy 
     
7

Note

The first example is clarified in the problem statement.

In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, 20 - - -, 25]$ $\to$ $[20, 15, 10, 15 - - -, 25]$ $\to$ $[20, 15, 10 - - -, 25]$ $\to$ $[20, 15 - - -, 25]$ $\to$ $[20 - - -, 25]$ $\to$ $[25]$ .

In the third example no bacteria can swallow any other bacteria.

#include<iostream>
#include<algorithm>
#include<string.h>
#include<set>
using namespace std;
#define ll long long
const int N = 1e6 + 5;
int a[N];
int main(int argc, char const *argv[]) {
    int n, k;
    scanf("%d %d", &n, &k);

    memset(a, 0, sizeof(a));
    set<int> ss;
    set<int> ::iterator it;
    set<int> ::iterator it_1;
    while(n--) {
        int x;
        scanf("%d", &x);
        ss.insert(x);
        a[x] ++;
    }
    it = ss.begin();
    for (; it != ss.end(); it ++) {
        int t = *(it) - k;
        it_1 = ss.lower_bound(t);
        while(it_1 != it) {
            //cout<<*(it_1)<<endl;
            ss.erase(it_1);
            it_1 = ss.lower_bound(t);
        }
    }
    it = ss.begin();
    int cnt = 0;
    for (; it != ss.end(); it ++) {
        cnt += a[*(it)];
    }
    printf("%d\n", cnt);
    return 0;
}

CF - B -Micro-World

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