HihoCoder-1403 suffix Array one • Repeat Melody

Description

Small hi usually a big hobby is playing the piano. We know that a musical melody is represented as a series of numbers consisting of N in length.

Little hi has practiced a lot of music and found that many of the works themselves contain the same melody. Melody is a continuous sequence of numbers, similar melodies can overlap in the original sequence. In 1 2 3 2 3 2 1, for example, 2 3 2 have occurred.

Little hi want to know what is the longest melody in a melody at least k times?

input

The first line is two integers n and K. 1≤n≤20000 1≤k≤n

Next there are N integers that represent the number of each tone. 1≤ Digital ≤100

Output

An integer line that represents the answer.

Sample Input

8 212323231

Sample Output

  4 
 
  
  > It is obvious that the maximum value of the minimum value of a continuous K-height array is asked,      
   " /strong>   Then there are two ways to do a binary or monotone queue 
 Here is a monotone queue    

#include <bits/stdc++.h>using namespace Std;const int maxn = 200005;int N, K, RK[MAXN], SA[MAXN], Y[MAXN], LN, p, M, C[maxn];int H[MAXN], WR[MAXN], Q[MAXN], L, R, A[maxn], POS[MAXN], k;bool cmp (int k1, int k2, int ls) {return WR[K1] = = WR [K2] && wr[k1 + ls] = = wr[k2 + ls];}  void SA (int x) {m = x;for (int i = 1; I <= n; ++i) c[rk[i]]++;for (int i = 1; I <= m; ++i) c[i] + = c[i-1];for (int i = n; I >= 1; i) sa[c[rk[i]]--] = I;ln = 1, p = 0;while (P < n) {k = 0;for (int i = n-ln + 1; I <= n; ++i) y[++k] = i;for (int i = 1; I <= N;  ++i) if (Sa[i] > ln) y[++k] = Sa[i]-Ln;memset (c, 0, sizeof c); for (int i = 1; I <= n; ++i) Wr[i] = rk[i];for (int i = 1; I <= N; ++i) c[wr[y[i]]]++; for (int i = 1; I <= m; ++i) c[i] + = c[i-1];for (int i = n; i >= 1; i.) sa[c[wr[y[i]]]--] = y[i];for (int i = 1; I <= N; ++i) Wr[i] = rk[i];rk[sa[1]] = 1; p = 1;for (int i = 2; I <= n; ++i) {if (!cmp (Sa[i], sa[i-1], LN)) ++p;rk[sa[i]] = p;}} ln <<=1; m = P;} void Gh () {k = 0;for (int i = 1; I <= n; ++i) rk[sa[i]] = i;for (int i = 1; I <= n; ++i) {if (k)--k; int J = Sa[rk[i ]-1];while (a[i + K] = = A[j + K]) ++k; H[rk[i]] = k;}} int main () {scanf ("%d%d", &n, &k), for (int i = 1; I <= n; ++i) scanf ("%d", &rk[i]), a[i] = Rk[i];sa ("G"); H (); int ans = 0;for (int i = 1; I <= n; ++i) {while (L < R && H[i] < Q[r]) r--;q[++r] = H[i]; Pos[r] = I;while (L < R && Pos[l] <= i-k + 1) ++l;if (i >= K) ans = max (ans, q[l]);} printf ("%d\n", ans); for (int i = 1; I <= n; ++i) cout << h[i] << ""; return 0;}

　　

HihoCoder-1403 suffix Array one • Repeat Melody