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)
Total Submission(s): 339    Accepted Submission(s): 165

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

 

题解：

前缀和真是个好东西。前缀和可以是前n个的和，前n项的最小值，前n项的最大值。。。

这个题用到前n项大于m的个数。然后用个尺取法。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = ;
int a[maxn];
int prek[maxn];
int main()
{
    int T;cin>>T;
    while(T--)
    {
        int n,m,k;
        cin>>n>>m>>k;
        memset(prek,,sizeof(prek));
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]>=m) prek[i] += prek[i-]+;
            else prek[i] = prek[i-];
        }
        long long ans = ;
        int sum = ;
        int t = ;
        for(int l=;l<=n;l++)
        {
            while(t<=n&&sum<k)
            {
                sum = prek[t]-prek[l];
                t++;
            }
            if(sum<k) break;
            ans += (n-t+);
            sum = prek[t-]-prek[l+];
        }
        printf("%I64d\n",ans);
    }
    return ;
}