HDU 1757 a simple math problem matrix fast power

If x <10 f (x) = x.
If X> = 10 f (x) = A0 * F (x-1) + A1 * F (X-2) + A2 * F (X-3) + ...... + A9 * F (X-10 );

For such a formula, you can easily construct such a matrix through the relationship between the matrix and linear transformation.

A0:

9
8
7
6
5
4
3
2
1
0

A1:
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0

Then f (n) = A1 ^ (N-9) * a0

Quick power

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <iostream>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 11;LL k, mod;struct Matrix {    int n, m;    LL data[maxn][maxn];    Matrix(int n = 0, int m = 0): n(n), m(m) {        memset(data, 0, sizeof(data));    }};Matrix operator * (Matrix a, Matrix b) {    Matrix ret(a.n, b.m);    for(int i = 1; i <= a.n; i++) {        for(int j = 1; j <= b.m; j++) {            for(int k = 1; k <= a.m; k++) {                ret.data[i][j] += a.data[i][k] * b.data[k][j];                ret.data[i][j] %= mod;            }        }    }    return ret;}Matrix pow(Matrix mat, LL p) {    if(p == 1) return mat;    Matrix ret = pow(mat * mat, p / 2);    if(p & 1) ret = ret * mat;    return ret;}int main() {    int a[10];    while(cin >> k >> mod) {        for(int i = 0; i < 10; i++) cin >> a[i];        Matrix A(10, 10), A0(10, 1);        if(k < 10) cout << k % mod << endl;        else {            for(int i = 1; i <= 10; i++) A.data[1][i] = a[i - 1];            for(int i = 2; i <= 10; i++) A.data[i][i - 1] = 1;            for(int i = 1, j = 9; i <= 10; i++, j--) A0.data[i][1] = j;            A = pow(A, k - 9);            A0 = A * A0;            cout << A0.data[1][1] << endl;        }    }    return 0;}

　　

 

HDU 1757 a simple math problem matrix fast power