zoukankan      html  css  js  c++  java
  • 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.

    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('
    ');
    }
     
    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<<"
    ";
            } 
            return 0;
    }
    

      

  • 相关阅读:
    小喵的在线共享编辑器
    简易漫画网站搭建-漫画喵Server版
    爬虫-漫画喵的100行逆袭
    应用OpenMP的一个简单的设计模式
    基于Caffe的Large Margin Softmax Loss的实现(中)
    基于Caffe的Large Margin Softmax Loss的实现(上)
    Oracle GoldenGate OGG管理员手册
    Spark快速大数据分析之RDD基础
    Apache Spark大数据分析入门(一)
    一文教你看懂大数据的技术生态圈:Hadoop,hive,spark
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5745171.html
Copyright © 2011-2022 走看看