【CF】323 Div2. D. Once Again...

挺有意思的一道题目。
考虑长度为n的数组，重复n次，可以得到n*n的最长上升子序列。同理，也可以得到n*n的最长下降子序列。
因此，把t分成prefix（上升子序列) + cycle(one integer repeating) + sufix(下降子序列)。
当t<=2*n时，暴力解。
注意数据范围。

 /* 583D */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int maxn = ;
 const int maxm = ;
 int c[maxm];
 int a[maxn], b[maxn*maxn*];
 int dp[maxm];
 int suf[maxn*maxn], pre[maxn*maxn];

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int n, t;
     int mx;
     int ans = ;

     scanf("%d %d", &n, &t);
     rep(i, , n+) {
         scanf("%d", &a[i]);
         ++c[a[i]];
     }

     if (t <= n*) {
         rep(i, , n+) {
             int k = i;
             rep(j, , t+) {
                 b[k] = a[i];
                 k += n;
             }
         }

         int n_ = n*t;

         memset(dp, , sizeof(dp));
         rep(i, , n_+) {
             mx = ;
             rep(j, , b[i]+)
                 mx = max(mx, dp[j]);
             pre[i] = ++mx;
             dp[b[i]] = mx;
         }
         rep(i, , n_+)
             ans = max(ans, pre[i]);
         printf("%d\n", ans);
         return ;
     }

     rep(i, , n+) {
         int k = i;
         rep(j, , n+) {
             b[k] = a[i];
             k += n;
         }
     }

     // calculate prefix
     int n_ = n*n;

     memset(dp, , sizeof(dp));
     rep(i, , n_+) {
         mx = ;
         rep(j, , b[i]+)
             mx = max(mx, dp[j]);
         pre[i] = ++mx;
         dp[b[i]] = mx;
     }

     // calculate suffix
     memset(dp, , sizeof(dp));
     per(i, , n_+) {
         mx = ;
         rep(j, b[i], maxm)
             mx = max(mx, dp[j]);
         suf[i] = ++mx;
         dp[b[i]] = mx;
     }

     // iteration
     int tmp, n2 = n+n;

     rep(i, , n+) {
         rep(j, , n+) {
             if (a[j] < a[i])
                 continue;
             tmp = pre[i+n_-n] + suf[j] + c[a[i]]*(t-n2);
             ans = max(ans, tmp);
         }
     }

     printf("%d\n", ans);

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }