bzoj2326 [HNOI2011]数学作业

bzoj2326 [HNOI2011]数学作业

求将 \(1\) 到 \(n\) 的所有整数顺序连接得到的数模 \(m\) 的值

\(n\leq10^{18},\ m\leq10^9\)

矩阵加速

---

首先得到 \(f_i=10^{\lfloor \log_{10}i\rfloor}\times f_{i-1}+i\)

形如 \(f_i=k\times f_{i-1}+i\) 的式子显然可以矩阵加速，即表示为 $$\begin{bmatrix}f_i&i&1\end{bmatrix}\begin{bmatrix}k&0&0\1&1&0\1&1&1\end{bmatrix}=\begin{bmatrix}f_{i+1}&i+1&1\end{bmatrix}$$

又因为 \(10^{\lfloor\log_{10}i\rfloor}\) 的取值只有 \(18\) 种，于是可以将每种取值分类讨论

即求出后 \(f_{10^a-1}\) 通过矩阵 \(\begin{bmatrix}10^{a+1}&0&0\\1&1&0\\1&1&1\end{bmatrix}\) 快速转移至 \(f_{10^{a+1}-1}\) ，多出来的部分同理计算即可

计算过程可能会爆 \(long\ long\) ……

时间复杂度 \(O(\log_{10}{n}\times\log_2n)\)

代码（码风极丑……

#include <bits/stdc++.h>
using namespace std;

#define rep(i) for (int i = 0; i < 3; i++)
typedef long long ll;
ll n; int P;

struct matrix {
  int arr[3][3];

  matrix() { memset(arr, 0, sizeof arr); }

  int* operator [] (const int& pos) {
    return arr[pos];
  }
} E, M;

inline int inc(int x, int y) {
  return x + y < P ? x + y : x + y - P;
}

matrix operator + (matrix a, matrix b) {
  rep(i) rep(j) a[i][j] = inc(a[i][j], b[i][j]);
  return a;
}

matrix operator * (matrix a, matrix b) {
  matrix res;
  rep(k) rep(i) rep(j) {
    res[i][j] = inc(res[i][j], 1ll * a[i][k] * b[k][j] % P);
  }
  return res;
}

matrix qp(matrix a, ll k) {
  matrix res = E;
  for (; k; k >>= 1, a = a * a) {
    if (k & 1) res = res * a;
  }
  return res;
}

int main() {
  E[0][0] = E[1][1] = E[2][2] = 1;
  M[1][0] = M[1][1] = M[2][0] = M[2][1] = M[2][2] = 1;
  scanf("%lld %d", &n, &P);
  int f[15];
  f[0] = 0;
  for (int i = 1; i < 10; i++) {
    f[i] = (10ll * f[i - 1] + i) % P;
  }
  if (n < 10) {
    printf("%d", f[n]);
    return 0;
  }
  matrix A;
  unsigned long long now = 10;
  A[0][0] = f[9], A[0][1] = 9, A[0][2] = 1;
  for (; now * 10 - 1 <= n; now *= 10) {
    M[0][0] = now * 10 % P;
    A = A * qp(M, now * 9);
  }
  M[0][0] = now * 10 % P;
  A = A * qp(M, n - now + 1);
  printf("%d", A[0][0]);
  return 0;
}