hdu - 1575 - Tr A（矩阵快速幂）

题意：求矩阵A的k次幂的主对角线上元素和模9973（(2 <= 矩（方）阵维数n <= 10)和k(2 <= k < 10^9)）。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1575

——>>矩阵快速幂。。。

#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 10 + 5;
const int mod = 9973;
int n, k;

struct Mar{
    int m[maxn][maxn];
    Mar(){
        memset(m, 0, sizeof(m));
    }
};

Mar operator + (Mar a, Mar b){
    Mar ret;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) ret.m[i][j] = (a.m[i][j] + b.m[i][j]) % mod;
    return ret;
}

Mar operator * (Mar a, Mar b){
    Mar ret;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            for(int l = 0; l < n; l++) ret.m[i][j] = (ret.m[i][j] + a.m[i][l] * b.m[l][j]) % mod;
    return ret;
}

Mar pow_mod(Mar a, int n){
    if(n == 1) return a;
    Mar x = pow_mod(a, n/2);
    x = x * x;
    if(n&1) x = x * a;
    return x;
}

void solve(){
    Mar A;
    int i, j, ret = 0;
    for(i = 0; i < n; i++)
        for(j = 0; j < n; j++) scanf("%d", &A.m[i][j]);
    A = pow_mod(A, k);
    for(i = 0; i < n; i++) ret = (ret + A.m[i][i]) % mod;
    printf("%d\n", ret);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &k);
        solve();
    }
    return 0;
}