HDU 1575 Tr A 矩阵快速幂

跟着cxlove的矩阵专题（http://blog.csdn.net/ACM_cxlove/article/details/7815594）刷的，一道一道来。

最裸的题目，直接快速幂算就好了。

#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_back
typedef 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 = 15;
const int mod = 9973;

struct Matrix {
    int n, m, data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }
    void print() {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                printf("%d ", data[i][j]);
            }
            puts("");
        }
    }
};

Matrix operator * (Matrix a, Matrix b) {
    int n = a.n, m = b.m;
    Matrix ret(n, m);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= 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, int k) {
    if(k == 1) return mat;
    Matrix ret = pow(mat * mat, k / 2);
    if(k & 1) ret = ret * mat;
    return ret;
}

Matrix mat;
int n, k;

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &k);
        mat.n = mat.m = n;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                scanf("%d", &mat.data[i][j]);
            }
        }
        mat = pow(mat, k);
        int ans = 0;
        for(int i = 1; i <= n; i++) {
            ans = (ans + mat.data[i][i]) % mod;
        }
        printf("%d\n", ans);
    }
    return 0;
}