hdu 5806 NanoApe Loves Sequence Ⅱ 前缀和+尺取法

NanoApeLovesSequenceⅡ算法解析

最新推荐文章于 2018-08-02 10:24:30 发布

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本文介绍了一道算法题目NanoApeLovesSequenceⅡ的解题思路及实现代码，该题目要求计算特定条件下连续子序列的数量。采用前缀和与两点法进行高效求解。

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

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;
}