hdu-5806 nanoape Loves Sequenceⅱ (ruler)

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Nanoape Loves Sequenceⅱ

Time limit:4000/2000 MS (java/others)

Memory limit:262144/131072 K (java/others)

Problem Descriptionnanoape, the retired Dog, have returned back to prepare for for the national Higher Education Entrance E xamination!

In the math class, Nanoape picked up sequences once again. He wrote down a sequence withNNumbers and a numbermOn the paper.

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

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

Inputthe first line of the input contains an integerT, denoting the number of test cases.

The first line of the input contains three integersn,m,k .

The second line of the input containsNIntegersa1,a2,.. . ,An , denoting the elements of the sequence.

1≤T≤10, 2≤N≤200000,1≤k ≤n / 2,  1≤m,ai≤< Span id= "mathjax-span-81" class= "Msubsup" >10 9

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

Sample Input17 4 24 2 7 7 6 5 1

Sample Output18 Test instructions: Give a sequence, ask the number of interval k greater than or equal to M of what is the number of intervals? Idea: Deal with the interval prefixes greater than or equal to M, and then the ruler method to engage; AC Code:

/************************************************┆┏┓┏┓┆┆┏┛┻━━━┛┻┓┆┆┃┃┆┆┃━┃┆┆┃┳┛┗┳┃┆┆┃┃ ┆┆┃┻┃┆┆┗━┓┏━┛┆┆┃┃┆┆┃┗━━━┓┆┆┃ac Horse ┣┓┆┆┃┏┛┆┆┗┓┓┏━┳┓┏┛┆┆┃┫┫┃┫┫┆ ┆┗┻┛┗┻┛┆************************************************ * * #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 Sequenceⅱ (ruler)