HDU 5806 NanoApe Loves Sequence Ⅱ [尺取法]

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http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=721&pid=1003
尺取法，先找k个>=m的，定位到r指针。然后移动l,如果a[l]>=m，说明现在区间内不足k个，r指针向后移，补到k个，如果a[l]

Code

#include <iostream>#include <cstdio>using namespace std;const int MAXN = 200000 + 9;int a[MAXN];void solve(){  int n, m, k;  scanf("%d%d%d", &n, &m, &k);  for (int i = 0; i < n; i++) {    scanf("%d", &a[i]);  }  long long ans = 0;  int r;  int t = 0;  for (r = 0; r < n; r++) {    if (a[r] >= m) t++;    if (t == k) break;  }  ans += (n - r);  for (int l = 1; l < n; l++) {    if (a[l - 1] >= m) {      t--;      r++;      for (;r < n; r++) {        if (a[r] >= m) t++;        if (t == k) break;      }      if (r >= n) break;      ans += (n - r);    } else {      ans += (n - r);    }  }  cout << ans << endl;}int main(){  //freopen("in", "r", stdin);  int t;  scanf("%d", &t);  while (t--) {    solve();  }}

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