Submission #697156

#TimeUsernameProblemLanguageResultExecution timeMemory
697156omikron123Osmosmjerka (COCI17_osmosmjerka)C++14
160 / 160
1758 ms79104 KiB
// https://oj.uz/problem/view/COCI17_osmosmjerka #include <cstdio> #include <algorithm> #include <functional> #include <vector> #include <cstring> #include <map> using namespace std; typedef long long ll; typedef pair<int,int> pi; // 从任一点出发,最多LCM(M,N)个字符之后,就会出现重复的pattern,因此我们只要生成这个长度的hash,就可以找到所有唯一字符串了。 // 然后出现重复的概率,就是每个unique pattern概率的平方和。 // // Hash要按binary jumping方式来算,这样就快了。 // // 因为数据较多(500*500*8=2e6),所以32位的hash是不够的,会有birthday paradox collision。办法是用两个32位hash,这个 // 就相当于64位,就可以了。 // // O(n^2logn) int m,n,K; // m,n <= 500, K <= 1e9 char s[505][505]; // M rows, N cols int dir[8][2] = {{0,1},{0,-1},{1,0},{-1,0},{1,1},{-1,-1},{1,-1},{-1,1}}; int hsh[2][23][505][505]; // hash value patterns for 23 log levels int p[2][23]; map<pi,int> cnts; // hash -> count const int A[2] = {911382323, 69420}, M[2] = {972663749, (int)1e9+9}; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int main() { scanf("%d %d %d", &m, &n, &K); for (int i = 0; i < m; i++) scanf("%s", s[i]); int lcm = m*n/gcd(m,n); if (K > lcm) K = lcm; for (int j = 0; j < 2; j++) { p[j][0] = A[j]; for (int i = 1; i < 23; i++) p[j][i] = (ll)p[j][i-1] * p[j][i-1] % M[j]; } for (int di = 0; di < 8; di++) { // 8 directions // calc hash values for step-len strings from each pos for (int lg = 0; lg < 23; lg++) { for (int r = 0; r < m; r++) for (int c = 0; c < n; c++) { if (lg == 0) hsh[0][di][r][c] = hsh[1][di][r][c] = s[r][c]; else { int R = (r + (1 << (lg-1)) * (m + dir[di][0])) % m; int C = (c + (1 << (lg-1)) * (n + dir[di][1])) % n; for (int j = 0; j < 2; j++) hsh[j][lg][r][c] = ((ll)hsh[j][lg-1][r][c] * p[j][lg-1] + hsh[j][lg-1][R][C]) % M[j]; } } } // calc hashes for K-len strings from each pos for (int r = 0; r < m; r++) for (int c = 0; c < n; c++) { int hs[2] = {0}; int R = r, C = c; for (int j = 0; j < 23; j++) { // binary jumping if ((K & (1 << j)) == 0) continue; for (int k = 0; k < 2; k++) hs[k] = ((ll)hs[k] * p[k][j] + hsh[k][j][R][C]) % M[k]; R = (R + (ll)(1<<j)*(m+dir[di][0])) % m; C = (C + (ll)(1<<j)*(n+dir[di][1])) % n; } cnts[{hs[0], hs[1]}]++; } } ll total = m*n*8; total = total*total; ll squared = 0; for (auto e: cnts) { int v = e.second; squared += (ll)v*v; } ll g = gcd(total, squared); printf("%lld/%lld", squared/g, total/g); return 0; }

Compilation message (stderr)

osmosmjerka.cpp: In function 'int main()':
osmosmjerka.cpp:36:10: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   36 |     scanf("%d %d %d", &m, &n, &K);
      |     ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
osmosmjerka.cpp:38:14: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   38 |         scanf("%s", s[i]);
      |         ~~~~~^~~~~~~~~~~~
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