BZOJ2326 [HNOI2011]数学作业

首先，列方程

我们定义s[i] = 10 ^ ((int) log(i))

于是，f[i] = (f[i - 1] * s[i] + i) % p

反正总之就是个沙茶递推

然后我们来看优化。。。怎么感觉像矩阵乘法呢？

发现要按照log(i)即i的位数分类讨论，在相同位数的时候令矩阵为

s[i] 0 0

1 1 0

0 1 1

即可

 1 /**************************************************************
 2     Problem: 2326
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:16 ms
 7     Memory:808 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11  
12 using namespace std;
13 typedef long long ll;
14  
15 ll n, p;
16  
17 struct Matrix {
18     ll x[4][4];
19      
20     Matrix (int X) {
21         int i, j;
22         for (i = 1; i <= 3; ++i)
23             for (j = 1; j <= 3; ++j)
24                 if (i == j) x[i][j] = X;
25                 else x[i][j] = 0;
26     }
27      
28     ll* operator [] (int X) {
29         return x[X];
30     }
31 };
32  
33 inline void operator *= (Matrix &x, Matrix y) {
34     int i, j, k;
35     Matrix res(0);
36     for (i = 1; i <= 3; ++i)
37         for (j = 1; j <= 3; ++j)
38             for (k = 1; k <= 3; ++k)
39                 (res[i][j] += x[i][k] * y[k][j]) %= p;
40     x = res;
41 };
42  
43 Matrix pow(Matrix x, ll y) {
44     Matrix res(1);
45     while (y) {
46         if (y & 1) res *= x;
47         x *= x, y >>= 1;
48     }
49     return res;
50 }
51  
52 int main() {
53     ll i;
54     Matrix ans(1), a(0);
55     scanf("%lld%lld", &n, &p);
56     a[1][1] = 10 % p, a[2][1] = a[2][2] = a[3][2] = a[3][3] = 1;
57     for (i = 10; i <= n; i *= 10, a[1][1] = i % p)
58         ans *= pow(a, i - i / 10);
59     ans *= pow(a, n - i / 10 + 1);
60     printf("%lld\n", (ans[2][1] + ans[3][1]) % p);
61     return 0;
62 }