hdu5806 NanoApe Loves Sequence Ⅱ 前缀和 + 二分

NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 662    Accepted Submission(s): 315

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

17 4 24 2 7 7 6 5 1

 

Sample Output

18

 

Source

BestCoder Round #86

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;ll n, m, k, a[200005], sum[200005], cnt, ans;int main(){ll T;cin >> T;while (T--) {scanf("%I64d%I64d%I64d", &n, &m, &k);sum[0] = a[0] = 0;//算到i为止（包括i）有多少个大于等于m的数//可知sum是不减的，可以用二分for (ll i = 1; i <= n; i++) {scanf("%I64d", &a[i]);sum[i] = (a[i] >= m ? sum[i - 1] + 1 : sum[i - 1]);}ans = 0;for (ll i = 1; i <= n; i++) {//core-debug l = i + k - 1 not l = i + 1 when k == 1ll l = i + k - 1, r = n, mid;while (l <= r) {mid = (l + r) / 2;ll t = sum[mid] - sum[i - 1];if (t >= k) {r = mid - 1;}else {l = mid + 1;}}if (l <= n) //二分找不到的情况，即所有都小ans += (n - l + 1);}printf("%I64d\n", ans);}return 0;}