A

HDU - 1575

很裸的矩阵乘法
对角线一共只有两条，试一下就好了

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

const int mod = 9973;

int n;

struct jz{
    int g[20][20];
    void init(){
        memset(g,0,sizeof(g));
    }
    void one(){
        memset(g,0,sizeof(g));
        for(int i = 1; i <= n; ++ i) g[i][i] = 1;
    }
};
jz operator * (jz a, jz b){
    jz c; c.init();
    for(int i = 1; i <= n; ++ i)
    for(int j = 1; j <= n; ++ j)
    for(int k = 1; k <= n; ++ k)
    c.g[i][j] += (a.g[i][k] * b.g[k][j]) % mod, c.g[i][j] %= mod;
    return c;
}
jz ksm(jz x,int y){
    jz z; z.one();
    while(y){
        if(y & 1) z = z * x;
        y >>= 1;
        x = x * x;
    }
    return z;
}

int k;
jz a;

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