【链接】 我是链接,点我呀:)
【题意】
在这里输入题意
【题解】
尺取法+二分。
类似滑动窗口。
即左端点为l,右端点为r.
维护a[r]-a[l]+1总是小于等于m的就好。
(大于m就右移左端点)
然后看看里面的数字个数是不是小于k;
不是的话让l..r中最右边那个数字删掉就好。
->链表优化一下即可。
【代码】
/*
1.Shoud it use long long ?
2.Have you ever test several sample(at least therr) yourself?
3.Can you promise that the solution is right? At least,the main ideal
4.use the puts("") or putchar() or printf and such things?
5.init the used array or any value?
6.use error MAX_VALUE?
7.use scanf instead of cin/cout?
8.whatch out the detail input require
*/
/*
sort(a+1,a+1+n);
L[i] = i-1;R[i] = i+1;
一开始l = 1,r = 0,now=0;
然后for (int i = 1;i <= n;i++){
r = i;
now++;
while(l <= r && a[r]-a[l]+1>m){
now--;
l=R[l];
}
if(now >=k){
for (int j = i,k = 1;k<=now-k+1;k++,j=L[j]){
delete(j);
ans++;
}
}
now = k-1;
}
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int n,m,k;
int a[N+10],L[N+10],R[N+10],ans;
void Delete(int pos){
}
int main(){
#ifdef LOCAL_DEFINE
freopen("rush_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(0),cin.tie(0);
cin >> n >> m >> k;
for (int i = 1;i <= n;i++) cin >> a[i];
sort(a+1,a+1+n);
for (int i = 1;i <= n;i++)
L[i] = i-1,R[i] = i+1;
R[0] = 1,L[n+1] = n;
int l = 1,r = 0,now = 0;
for (int i = 1;i <= n;i++){
r = i;
now++;
while(l <= r && a[r]-a[l]+1>m){
if (a[l]!=0) now--;
l=R[l];
}
if(now >=k){
a[r] = 0;
int ll = L[r],rr = R[r];
R[ll] = rr;
L[rr] = ll;
now--;
ans++;
}
}
cout << ans << endl;
return 0;
}