Submission #587052

#TimeUsernameProblemLanguageResultExecution timeMemory
587052eecsLjetopica (COI19_ljetopica)C++17
100 / 100
64 ms460 KiB
#include <bits/stdc++.h> using namespace std; const int maxn = 1010, P = 1000000007; int n, K, pw[maxn]; string dir, A, B; typedef array<int, 2> arr; arr f[maxn][2][2][2], g[maxn][2][2][2]; void operator += (arr &A, const arr &B) { A = {(A[0] + B[0]) % P, (A[1] + B[1]) % P}; } arr operator * (const arr &A, const arr &B) { return {1LL * A[0] * B[0] % P, (1LL * A[0] * B[1] + 1LL * A[1] * B[0]) % P}; } int main() { ios::sync_with_stdio(0), cin.tie(0); cin >> n >> K; for (int i = pw[0] = 1; i < n; i++) { pw[i] = 2 * pw[i - 1] % P; } cin >> dir >> A >> B; f[0][0][0][0] = f[0][1][0][0] = {1, pw[n - 1]}; for (int i = 0; i < n - 1; i++) { memset(g, 0, sizeof(g)); for (int j = 0; j <= K; j++) for (int k : {0, 1}) { for (int a : {0, 1}) for (int b : {0, 1}) if (f[j][k][a][b][0]) { for (int l : {0, 1}) { char c = (dir[i] == 'L') ^ k ^ l ? '0' : '1'; if (!a && c < A[i + 1]) continue; if (!b && c > B[i + 1]) continue; g[j + l][k ^ l][a | (c > A[i + 1])][b | (c < B[i + 1])] += f[j][k][a][b] * arr{1, (c - '0') * pw[n - i - 2]}; } } } swap(f, g); } int ans = 0; for (int i : {0, 1}) for (int j : {0, 1}) for (int k : {0, 1}) { (ans += f[K][i][j][k][1]) %= P; } cout << ans << "\n"; return 0; }

Compilation message (stderr)

ljetopica.cpp: In function 'arr operator*(const arr&, const arr&)':
ljetopica.cpp:15:31: warning: narrowing conversion of '(((1 * ((long long int)((int)(& A)->std::array<int, 2>::operator[](0)))) * ((long long int)((int)(& B)->std::array<int, 2>::operator[](0)))) % ((long long int)((int)P)))' from 'long long int' to 'int' [-Wnarrowing]
   15 |     return {1LL * A[0] * B[0] % P, (1LL * A[0] * B[1] + 1LL * A[1] * B[0]) % P};
      |             ~~~~~~~~~~~~~~~~~~^~~
ljetopica.cpp:15:76: warning: narrowing conversion of '((((1 * ((long long int)((int)(& A)->std::array<int, 2>::operator[](0)))) * ((long long int)((int)(& B)->std::array<int, 2>::operator[](1)))) + ((1 * ((long long int)((int)(& A)->std::array<int, 2>::operator[](1)))) * ((long long int)((int)(& B)->std::array<int, 2>::operator[](0))))) % ((long long int)((int)P)))' from 'long long int' to 'int' [-Wnarrowing]
   15 |     return {1LL * A[0] * B[0] % P, (1LL * A[0] * B[1] + 1LL * A[1] * B[0]) % P};
      |                                    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~
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