HDU 5806 NanoApe Loves Sequence Ⅱ

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

给一个序列,求 第K大的数大于等于m的区间 的个数是多少？ 尺取法  

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #include <vector> #include <cmath> #include <stack> #include <string> #include <map> #include <set> #define pi acos(-1) #define LL long long #define ULL unsigned long long #define inf 0x3f3f3f3f #define INF 1e18 #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 using namespace std; typedef pair<int, int> P; const int maxn = 5e5 + 5; LL a[maxn]; int main(void) { // freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin); LL T, i, j, n, m, k, cur, ans, cnt; scanf("%I64d", &T); while (T--) { scanf("%I64d %I64d %I64d", &n, &m, &k); for (i = 1; i <= n; i++) scanf("%I64d", &a[i]); cur = 1; ans = cnt = 0; for (i = 1; i <= n; i++){ if (a[i] >= m) cnt++; if (cnt >= k){ while (cnt >= k){ ans += n - i + 1; if (a[cur] >= m) cnt--; cur++; } } } printf("%I64d\n", ans); } return 0; }