HDU5806 NanoApe Loves Sequence Ⅱ

最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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文章标签： java

原文链接：http://www.cnblogs.com/pk28/p/5745762.html

本文介绍了一个关于序列处理的问题：如何找出所有连续子序列中，使得子序列中第k大的数不小于给定值m的方法。通过将符合条件的数标记为1，其余标记为0，利用双指针法优化计算过程。

NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 517 Accepted Submission(s): 250

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

Input

The first line of the input contains an integer

Output

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

Sample Input

7 4 2

4 2 7 7 6 5 1

Sample Output

求有多少个区间，使得区间内的第k大的值>=m.

将不小于m的数看作1，剩下的数看作0，那么只要区间内1的个数不小于k则可行，枚举左端点，右端点可以通过two-pointer求出。

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/7 10:02:54
File Name     :hdu5806.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int a[200010];
int sum[200010];
int n,m,k;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    while(t--){
        scanf("%d%d%d",&n,&m,&k);
        int x;
        sum[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(x>=m)a[i]=1;
            else a[i]=0;
            sum[i]=sum[i-1]+a[i];
        }
        ll ans=0;
        int r=1;
        for(int l=1;l<=n;l++){//枚举左端点
            while(r<=n&&sum[r]-sum[l-1]<k)r++;
            if(r>n)break;
            ans+=(n-r+1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}