[HDOJ6395] Sequence (反演思想，整除分块，矩阵快速幂)

发表于 | 分类于 ， ， | 评论数： | 阅读次数：

链接：http://acm.hdu.edu.cn/showproblem.php?pid=6395

就算个式子。

这题关键在于如何处理后面的下取整分式，会发现如何构造这个矩阵都不太好弄这个式子。知道一点莫比乌斯反演，考虑到了按照整除结果对取整式进行分块，每一块单独跑块长次矩阵快速幂，然后累计到结果上就行。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL maxn = 5;
const LL mod = 1e9+7;
typedef struct Matrix {
  LL m[maxn][maxn];
  LL r;
  LL c;
  Matrix(){
    r = c = 0;
    memset(m, 0, sizeof(m));
  } 
} Matrix;
Matrix mul(Matrix m1, Matrix m2) {
  Matrix ret = Matrix();
  ret.r = m1.r; ret.c = m2.c;
  for(LL i = 1; i <= m1.r; i++) {
    for(LL j = 1; j <= m2.r; j++) {
      for(LL k = 1; k <= m2.c; k++) {
        if(m2.m[j][k] == 0) continue;
        ret.m[i][k] = (ret.m[i][k] + (m1.m[i][j] * m2.m[j][k]) % mod) % mod;
      }
    }
  }
  return ret;
}
Matrix quickmul(Matrix m, LL n) {
  Matrix ret = Matrix();
  for(LL i = 1; i <= m.r; i++) ret.m[i][i]  = 1;
  ret.r = m.r;
  ret.c = m.c;
  while(n) {
    if(n & 1) ret = mul(m, ret);
    m = mul(m, m);
    n >>= 1;
  }
  return ret;
}

LL a, b, c, d, p, n;

signed main() {
  // freopen("in", "r", stdin);
  LL T;
  scanf("%lld", &T);
  Matrix mat, ret;
  while(T--) {
    scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&d,&p,&n);
    mat.r = mat.c = 3;
    ret.r = 3; ret.c = 1;
    mat.m[1][1] = d, mat.m[1][2] = c;
    mat.m[2][1] = 1; mat.m[2][2] = 0; mat.m[2][3] = 0;
    mat.m[3][1] = 0; mat.m[3][2] = 0; mat.m[3][3] = 1;
    ret.m[1][1] = b; ret.m[2][1] = a; ret.m[3][1] = 1;
    if(n == 1) {
      printf("%lld\n", a);
      continue;
    }
    if(n < p) {
      for(LL i = 3; i <= n; i=p/(p/i)+1) {
        mat.m[1][1] = d, mat.m[1][2] = c;
        mat.m[2][1] = 1; mat.m[2][2] = 0; mat.m[2][3] = 0;
        mat.m[3][1] = 0; mat.m[3][2] = 0; mat.m[3][3] = 1;
        mat.m[1][3] = (p / i);
        if(n <= p / (p/i)) mat = quickmul(mat, n-i+1);
        else mat = quickmul(mat, p/(p/i)+1-i);
        ret = mul(mat, ret);
      }
      printf("%lld\n", ret.m[1][1]);
      continue;
    }
    for(LL i = 3; i <= p; i=p/(p/i)+1) {
      mat.m[1][1] = d, mat.m[1][2] = c;
      mat.m[2][1] = 1; mat.m[2][2] = 0; mat.m[2][3] = 0;
      mat.m[3][1] = 0; mat.m[3][2] = 0; mat.m[3][3] = 1;
      mat.m[1][3] = (p / i);
      mat = quickmul(mat, p/(p/i)+1-i);
      ret = mul(mat, ret);
    }
    mat.m[1][1] = d, mat.m[1][2] = c;
    mat.m[2][1] = 1; mat.m[2][2] = 0; mat.m[2][3] = 0;
    mat.m[3][1] = 0; mat.m[3][2] = 0; mat.m[3][3] = 1;
    mat.m[1][3] = 0LL;
    mat = quickmul(mat, n-max(p, 2LL));
    ret = mul(mat, ret);
    printf("%lld\n", ret.m[1][1]);
  }
  return 0;
}