hdu-5806 NanoApe Loves Sequence Ⅱ(尺取法)

题目链接：

NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others)   

 Memory Limit: 262144/131072 K (Java/Others)

Problem Description

NanoApe, the Retired Dog, has returned back to prepare for for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with n numbers and a number m on the paper.

Now he wants to know the number of continous subsequences of the sequence in such a manner that the k-th largest number in the subsequence is no less than m.

Note : The length of the subsequence must be no less than k.

 

Input

The first line of the input contains an integer T, denoting the number of test cases.

In each test case, the first line of the input contains three integers n,m,k.

The second line of the input contains n integers A1,A2,...,An, denoting the elements of the sequence.

1≤T≤10, 2≤n≤200000, 1≤k≤n/2, 1≤m,Ai≤109

 

Output

For each test case, print a line with one integer, denoting the answer.

 

Sample Input

1
7 4 2
4 2 7 7 6 5 1

 

Sample Output

18

 

题意：

 

给一个序列,问区间第K大的数大于等于m的区间个数是多少？

 

思路：

 

处理出大于等于m的区间前缀和,然后尺取法搞;

 

AC代码：

 

/************************************************
┆  ┏┓　　　┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　┻　　　┃ ┆
┆┗━┓　  　┏━┛ ┆
┆　　┃　  　┃　　┆　　　　　　
┆　　┃　  　┗━━━┓ ┆
┆　　┃　 AC代马 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */ 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e5+10;
const int maxn=2e3+14;
const double eps=1e-12;

int n,m,a[N],sum[N],k;

int main()
{
        int t;
        read(t);
        while(t--)
        {
            read(n);read(m);read(k);
            For(i,1,n)
            {
                read(a[i]);
                if(a[i]>=m)sum[i]=sum[i-1]+1;
                else sum[i]=sum[i-1];
            }
            LL ans=0;
            int r=1;
            For(i,1,n-k+1)
            {
                r=max(i+k-1,r);
                while(sum[r]-sum[i-1]<k&&r<=n)r++;
                if(r<=n&&r-i+1>=k)ans=ans+(n-r+1);
            }
            cout<<ans<<"\n";
        }
        return 0;
}