hdu 5806 NanoApe Loves Sequence Ⅱ 前缀和+尺取法

NanoApe Loves Sequence Ⅱ,原文链接,click here

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 596 Accepted Submission(s): 286

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

Source

BestCoder Round #86

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前缀和+两点法.

#include<stdio.h>#include<iostream>#include<algorithm>#include<string>#include<string.h>using namespace std;int f[200005];int main(){    int t,n,m,k,a;    scanf("%d",&t);    while(t--)    {        f[0]=0;        memset(f,0,sizeof(f));        scanf("%d%d%d",&n,&m,&k);        for(int i=1;i<=n;i++)        {            scanf("%d",&a);            f[i]=f[i-1] + (a>=m?1:0);        }        long long ans=0;        int j=1;        for(int i=1;j<=n;)        {            if(f[j]-f[i-1]>=k)            {                ans+=n-j+1;                i++;            }            else            {                if(j<=n)                    j++;            }        }        printf("%lld\n",ans);    }    return 0;}