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

/**************************************************************
    Problem: 2326
    User: rausen
    Language: C++
    Result: Accepted
    Time:16 ms
    Memory:808 kb
****************************************************************/
 
#include <cstdio>
 
using namespace std;
typedef long long ll;
 
ll n, p;
 
struct Matrix {
    ll x[4][4];
     
    Matrix (int X) {
        int i, j;
        for (i = 1; i <= 3; ++i)
            for (j = 1; j <= 3; ++j)
                if (i == j) x[i][j] = X;
                else x[i][j] = 0;
    }
     
    ll* operator [] (int X) {
        return x[X];
    }
};
 
inline void operator *= (Matrix &x, Matrix y) {
    int i, j, k;
    Matrix res(0);
    for (i = 1; i <= 3; ++i)
        for (j = 1; j <= 3; ++j)
            for (k = 1; k <= 3; ++k)
                (res[i][j] += x[i][k] * y[k][j]) %= p;
    x = res;
};
 
Matrix pow(Matrix x, ll y) {
    Matrix res(1);
    while (y) {
        if (y & 1) res *= x;
        x *= x, y >>= 1;
    }
    return res;
}
 
int main() {
    ll i;
    Matrix ans(1), a(0);
    scanf("%lld%lld", &n, &p);
    a[1][1] = 10 % p, a[2][1] = a[2][2] = a[3][2] = a[3][3] = 1;
    for (i = 10; i <= n; i *= 10, a[1][1] = i % p)
        ans *= pow(a, i - i / 10);
    ans *= pow(a, n - i / 10 + 1);
    printf("%lld
", (ans[2][1] + ans[3][1]) % p);
    return 0;
}