做防腐木花架的网站/怎么优化网络

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1575

题意：求矩阵的k次方后主对角线上的和+mod9973

解法：矩阵快速幂的入门题。

总结:感觉矩阵快速幂还挺有趣的，今天开始了解了一下。加油！

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
int n,k;
struct matrix
{
    int an[20][20];
};
matrix temp;
matrix multiply(matrix a,matrix b)
{
    matrix c;
    memset(c.an,0,sizeof(c.an));
    for (int i=1 ;i<=n ;i++)
    {
        for (int j=1 ;j<=n ;j++)
        {
            for (int k=1 ;k<=n ;k++)
            {
                c.an[i][j] += (a.an[i][k]*b.an[k][j])%9973;
                c.an[i][j] %= 9973;
            }
        }
    }
    return c ;
}
int calc(int u)
{
    matrix x;
    memset(x.an,0,sizeof(x.an));
    for (int i=1 ;i<=n ;i++) x.an[i][i]=1;
    while (u)
    {
        if (u & 1) x=multiply(x,temp);
        u >>= 1;
        temp=multiply(temp,temp);
    }
    int sum=0;
    for (int i=1 ;i<=n ;i++)
    {
        sum += (x.an[i][i])%9973;
        sum %= 9973 ;
    }
    return sum%9973;
}
int main()
{
    int t;
    cin>>t;
    while (t--)
    {
        cin>>n>>k;
        for (int i=1 ;i<=n ;i++)
        {
            for (int j=1 ;j<=n ;j++)
            scanf("%d",&temp.an[i][j]);
        }
        printf("%d\n",calc(k));
    }
    return 0;
}