B. Prison Transfer(RMQ入门，线段树入门）

B. Prison Transfer

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The prison of your city has ​n​ prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer ​c​ of the prisoners to a prison located in another city.

For this reason, he made the ​n​ prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.

Then, the mayor told you to choose the ​c​ prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,

• • 
    
• The chosen ​c​ prisoners has to form a contiguous segment of prisoners.
• Any of the chosen prisoner's crime level should not be greater then ​t​. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
  • 
    

Find the number of ways you can choose the ​c​ prisoners.

Input

The first line of input will contain three space separated integers ​n​ (1 ≤ ​n​ ≤ 2·105), ​t​ (0 ≤ ​t​ ≤ 109) and ​c​ (1 ≤ ​c​ ≤ ​n​). The next line will contain ​n​ space separated integers, the ​ith​ integer is the severity ​ith​ prisoner's crime. The value of crime severities will be non-negative and will not exceed 109.

Output

Print a single integer — the number of ways you can choose the ​c​ prisoners.

Examples

input

Copy

4 3 3
2 3 1 1

output

Copy

2

input

Copy

1 1 1
2

output

Copy

0

input

Copy

11 4 2
2 2 0 7 3 2 2 4 9 1 4

output

Copy

6

这题大概意思就是，给n t c，在n个数据中连续c个最大值小于t的个数，这题用RMQ可以 做，很明显线段树也是可以做的。RMQ：

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;

int n,t,c;
int ans;

int RMQ_max[200010][20];
double Limit;
void init(int n){
  Limit = log(n)/log(2.0);
  int tmp;
  for(int i=1;i<=n;i++){
    scanf("%d",&tmp);
    RMQ_max[i][0] = tmp;
  }
  for(int i=1;i<=Limit;i++){
    for(int j=1;j+(1<<i)-1<=n;j++){
      RMQ_max[j][i] = max(RMQ_max[j][i-1],RMQ_max[j+(1<<(i-1))][i-1]);
    }
  }
}

void RMQ_find(){
  for(int i=1;i<=n-c+1;i++){
    int l = i;
    int r = i+c-1;
    int k = (int)(log(r-l+1)/log(2.0));
    int MAXN = max(RMQ_max[l][k],RMQ_max[r-(1<<k)+1][k]);
    if(MAXN <= t) ans++;
  }
}
int main(){
  while(~scanf("%d%d%d",&n,&t,&c)){
    init(n);
    ans = 0;
    RMQ_find();
    printf("%d\n",ans);
  }
  return 0;
}

//2 2 0 7 3 2 2 4 9 1 4

线段树做法：

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
int mx[maxn*4];
int n,t,c;
void build(int l,int r,int id)
{
  if(l==r)
  {
    int x;
    scanf("%d",&x);
    mx[id]=x;
    return ;
  }
  int mid=(l+r)>>1;
  build(l,mid,id<<1);
  build(mid+1,r,id<<1|1);
  mx[id]=max(mx[id<<1],mx[id<<1|1]);
}
int qu(int l,int r,int ql,int qr,int id)
{
  if(ql<=l&&r<=qr)
  {
    return mx[id];
  }
  int mid=(l+r)>>1;
  int ans=0;
  if(ql<=mid) ans=max(ans,qu(l,mid,ql,qr,id<<1));
  if(qr>mid) ans=max(ans,qu(mid+1,r,ql,qr,id<<1|1));
  return ans;
}
int main()
{
  scanf("%d%d%d",&n,&t,&c);
  build(1,n,1);
  int sum=0;
  for(int i=1;i+c-1<=n;i++)
  {
    int ans=qu(1,n,i,i+c-1,1);
  //  printf("%d\n",ans);
    if(ans<=t) sum++;
  }
  printf("%d",sum);
}