Raslboyy

Codeforces F1 ТехноКубок 3

Dec 11th, 2019
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C++ 2.13 KB | None | 0 0
  1. #include <iostream>
  2. #include <vector>
  3. #include <list>
  4. #include <algorithm>
  5. #include <string>
  6. #include <math.h>
  7. #include <climits>
  8.  
  9. //#include <chrono>
  10. //#include <fstream>
  11. //#include <bits/stdc++.h>
  12.  
  13. #define mp make_pair
  14. #define pb push_back
  15. #define eb emplace_back
  16. #define x first
  17. #define y second
  18. #define sz(x) (int)x.size()
  19. #define all(x) begin(x), end(x)
  20. #define rall(x) rbegin(x), rend(x)
  21. #define FOR(i,a,b) for (int i = (a); i < (b); i++)
  22. #define RFOR(i,b,a) for (int i = (b) - 1; i >= (a); i--)
  23.  
  24. using namespace std;
  25.  
  26. typedef unsigned long long ull;
  27. typedef long long ll;
  28. typedef long double ld;
  29. typedef pair<int, int> pi;
  30. typedef pair<ll, ll> pl;
  31. typedef vector<int> veci;
  32. typedef vector<bool> vecb;
  33. typedef vector<vector<int>> vvi;
  34. typedef vector<vector<bool>> vvb;
  35.  
  36. const int INF_I = 1e9;
  37. const ll INF_LL = 1e18;
  38.  
  39. const int MOD = 998244353;
  40.  
  41.  
  42. int main() {
  43.     /*ifstream cin("in.txt");
  44.     ofstream cout("out.txt");
  45.     auto start_time = chrono::high_resolution_clock::now();*/
  46.  
  47.     ios::sync_with_stdio(false);
  48.     cin.tie(0);
  49.     cout.tie(0);
  50.  
  51.    
  52.     int n, k;
  53.     cin >> n >> k;
  54.     const int ST = n;
  55.     veci vec1(n), vec2(n);
  56.     FOR(i, 0, n) {
  57.         cin >> vec1[i];
  58.         vec2[(i + 1) % n] = vec1[i];
  59.     }
  60.     /*swap(vec2[0], vec2[n]);*/
  61.     vvi dp(n + 1, veci(ST + n + 10));
  62.     /*FOR(i, 0, n + 1)
  63.         dp[0][i] = 1;*/
  64.     dp[0][ST] = 1;
  65.     for (int i = 0; i < n; i++) {
  66.         for (int j = 0; j < ST + n + 10; j++) {
  67.             if (vec1[i] == vec2[i]) {
  68.                 dp[i + 1][j] = (dp[i + 1][j] + ((ll)dp[i][j] * k) % MOD) % MOD;
  69.             }
  70.             else {
  71.                 if (j + 1 <= ST + n + 2)
  72.                     dp[i + 1][j+1] = (dp[i + 1][j + 1] + ((ll)dp[i][j]) % MOD) % MOD;
  73.                 if (j - 1 >= 0)
  74.                     dp[i + 1][j - 1] = (dp[i + 1][j - 1] + ((ll)dp[i][j]) % MOD) % MOD;
  75.                 if (k > 1)
  76.                     dp[i + 1][j] = (dp[i + 1][j] + ((ll)dp[i][j] * (k - 2)) % MOD) % MOD;
  77.             }
  78.         }
  79.     }
  80.  
  81.     int answ = 0;
  82.     FOR(i, ST + 1, ST + n + 2)
  83.         answ = (answ + (ll)dp[n][i]) % MOD;
  84.     cout << answ << endl;
  85.    
  86.     /*auto end_time = chrono::high_resolution_clock::now();
  87.     chrono::duration<double> duration = end_time - start_time;
  88.     cout << duration.count() << "\n";
  89.     cin.close();
  90.     cout.close();*/
  91.     return 0;
  92. }
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