[HIHO1560] H国的身份证号码II（dp，计数，矩阵快速幂）

题目链接：http://hihocoder.com/problemset/problem/1560

有了数位dp的思维基础，这个递推式很容易想到：

构造转移矩阵的时候思考数位dp的转移。

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

typedef long long LL;
const LL mod = 1e9+7;
const int maxn = 11;

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

LL n;
int k;

signed main() {
    // freopen("in", "r", stdin);
    while(~scanf("%lld%d",&n,&k)) {
        Matrix p; p.r = p.c = 10;
        for(int i = 0; i <= min(k, 9); i++) {
            for(int j = 0; j <= min(k, 9); j++) {
                if(i * j <= k) p.m[i][j] = 1;
            }
        }
        p = quickmul(p, n-1LL, mod);
        LL ret = 0;
        for(int i = 0; i <= min(k, 9); i++) {
            for(int j = 1; j <= min(k, 9); j++) {
                ret += p.m[i][j];
                ret %= mod;
            }
        }
        printf("%lld
", ret);
    }
    return 0;
}