HDU 1575 Tr A (矩阵快速幂)

大意：求出矩阵的对角线元素的和。

思路：裸的巨快，最后循环求和。

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1

int n ;
const int mod = 9973;
struct node{
    int r[15][15];
}q;

void matrix_pow(node &a,node &b){
    int i,j,k;
    node t;
    memset(t.r,0,sizeof(t.r));
    for(k = 0;k < n;++ k){
        for(i =0 ;i < n;i++){
            for(j = 0;j < n;++ j){
                t.r[i][j] = (t.r[i][j]+ (a.r[i][k]*b.r[k][j]))%mod;
            }
        }
    }
    a = t;
}

node so(node &q,int m){
    int i,j,k;
    node tmp;
    memset(tmp.r,0,sizeof(tmp.r));
    for(i=0;i<n;++i)
        tmp.r[i][i]=1;
    while(m){
        if(m&1){
            matrix_pow(tmp,q);
        }

        matrix_pow(q,q);
        m =m >> 1;
    }
    q = tmp;
}

int main(){
    int k,i,j,cla;
    scanf("%d",&cla);
    while(cla--){
        scanf("%d%d",&n,&k);
        memset(q.r,0,sizeof(q.r));
        for(i = 0;i < n; ++ i)
            for(j = 0;j < n ;++ j)
                scanf("%d",&q.r[i][j]);
        so(q,k);
        __int64 ans = 0;
        for(i = 0;i < n;++ i){
            ans = (ans+q.r[i][i])%mod;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}