LibreOJ #2002. 「SDOI2017」序列计数
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/* LibreOJ #2002. 「SDOI2017」序列计数 线性筛 + 矩阵优化dp 先构造出全部情况的矩阵 用矩阵快速幂计算答案 再构造出全不是质数的矩阵 计算出答案 前一个答案减后一个答案即可 */ #include <cstdio> #include <iostream> #include <cstring> const int BUF = 12312312; char Buf[BUF], *buf = Buf; #define Mod 20170408 inline void read (int &now) { for (now = 0; !isdigit (*buf); ++ buf); for (; isdigit (*buf); now = now * 10 + *buf - ‘0‘, ++ buf); } #define Max 20000007 #define L 100 struct Matrix { int c[L][L]; Matrix () { memset (c, 0, sizeof c); } Matrix operator * (const Matrix &now) const { Matrix res; for (register int i = 0, j, k; i < L; ++ i) for (k = 0; k < L; ++ k) if (this->c[i][k]) for (j = 0; j < L; ++ j) res.c[i][j] = (res.c[i][j] + 1LL * this->c[i][k] * now.c[k][j]) % Mod; return res; } }; int operator ^ (Matrix A, int p) { Matrix res; res.c[0][0] = 1; for (; p; A = A * A, p >>= 1) if (p & 1) res = res * A; return res.c[0][0]; } Matrix A; int prime[Max]; bool is[Max]; int C; void Euler (const int N) { is[1] = true; for (register int i = 2, j; i <= N; ++ i) { if (!is[i]) prime[++ C] = i; for (j = 1; i * prime[j] <= N && j <= C; ++ j) { is[i * prime[j]] = true; if (i % prime[j] == 0) break; } } } int data[Max]; int Main () { fread (buf, 1, BUF, stdin); int N, M, K; read (N), read (M), read (K); Euler (Max - 7); register int i, j; for (i = 1; i <= M; ++ i) ++ data[i % K]; for (i = 0; i < K; ++ i) for (j = 0; j < K; ++ j) A.c[i][(i + j) % K] = data[j] % Mod; int Answer = A ^ N; memset (A.c, 0, sizeof A.c); memset (data, 0, sizeof data); for (i = 1; i <= M; ++ i) if (is[i]) ++ data[i % K]; for (i = 0; i < K; ++ i) for (j = 0; j < K; ++ j) A.c[i][(i + j) % K] = data[j] % Mod; Answer -= A ^ N; printf ("%d", (Answer + Mod) % Mod); return 0; } int ZlycerQan = Main (); int main (int argc, char *argv[]) {;}
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