UVA 11149 Power of Matrix

题目链接：UVA 11149 Power of Matrix

题目大意：
已知\(n\)阶矩阵\(A\)，求\(\sum_{i=1}^{k}A^i\)。

题解：
定义\({SUM}_i=\sum_{j=1}^{i}A^j\)。
构造矩阵：

\[\left( \begin{matrix} A^i \\ {SUM}_i \end{matrix} \right) = \left( \begin{matrix} A & 0 \\ A & E \end{matrix} \right) \times \left( \begin{matrix} A^{i-1} \\ {SUM}_{i-1} \end{matrix} \right) \]

所以：

\[\left( \begin{matrix} A^k \\ {SUM}_k \end{matrix} \right) = \left( \begin{matrix} A & 0 \\ A & E \end{matrix} \right)^{k-1} \times \left( \begin{matrix} A \\ A \end{matrix} \right) \]

题目只要求最后一位数字，则在矩阵快速幂时对\(10\)取余就行了。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

struct Matrix {  // 矩阵
    int row, col;
    int num[85][85];
};

Matrix multiply(Matrix a, Matrix b) {  // 矩阵乘法
    Matrix temp;
    temp.row = a.row, temp.col = b.col;
    memset(temp.num, 0, sizeof(temp.num));
    for (int i = 0; i < a.row; ++i)
        for (int j = 0; j < b.col; ++j)
            for (int k = 0; k < a.col; ++k)
                temp.num[i][j] = (temp.num[i][j] + a.num[i][k] * b.num[k][j]) % 10;
    return temp;
}

Matrix MatrixFastPow(Matrix base, int n, int k) {  // 矩阵快速幂
    Matrix ans;
    ans.row = ans.col = n;
    memset(ans.num, 0, sizeof(ans));
    for (int i = 0; i < n; ++i) {
        ans.num[i][i] = 1;
    }
    while (k) {
        if (k & 1) ans = multiply(ans, base);
        base = multiply(base, base);
        k >>= 1;
    }
    return ans;
}

int main() {
    int n, k;
    while (cin >> n >> k && n) {
        Matrix base, res;
        base.row = base.col = 2 * n;
        res.row = 2 * n, res.col = n;
        memset(base.num, 0, sizeof(base.num));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                cin >> base.num[i][j];
                base.num[i][j] %= 10; // 只需要最后一位
                res.num[i][j] = res.num[i + n][j] = base.num[i + n][j] = base.num[i][j];
            }
        }
        for (int i = n; i < 2 * n; ++i) {
            base.num[i][i] = 1;
        }
        Matrix ans = multiply(MatrixFastPow(base, 2 * n, k - 1), res);
        for (int i = n; i < 2 * n; ++i) {
            for (int j = 0; j < n; ++j) {
                cout << ans.num[i][j];
                if (j != n - 1) cout << ' ';
            }
            cout << endl;
        }
        cout << endl;
    }
    return 0;
}