码迷,mamicode.com
首页 > 其他好文 > 详细

hdu-5806 NanoApe Loves Sequence Ⅱ(尺取法)

时间:2016-08-07 00:40:49      阅读:165      评论:0      收藏:0      [点我收藏+]

标签:

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.

1T10, 2n200000, 1kn/2, 1m,Ai109
 

 

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;
}

  

hdu-5806 NanoApe Loves Sequence Ⅱ(尺取法)

标签:

原文地址:http://www.cnblogs.com/zhangchengc919/p/5745171.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!