hdu 1575 Tr A（矩阵乘法）

Tr A

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4187    Accepted Submission(s): 3128

Problem Description

A为一个方阵，则Tr A表示A的迹（就是主对角线上各项的和），现要求Tr(A^k)%9973。

 

Input

数据的第一行是一个T，表示有T组数据。
每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行，每行有n个数据，每个数据的范围是[0,9]，表示方阵A的内容。

 

Output

对应每组数据，输出Tr(A^k)%9973。

 

Sample Input

  
  

   
   2
2 2
1 0
0 1
3 99999999
1 2 3
4 5 6
7 8 9
  
  

 

Sample Output

  
  

   
   2
2686
  
  

 

Author

xhd

 

Source

HDU 2007-1 Programming Contest

 

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define p 9973
using namespace std;
struct data{
  int	a[20][20];
}a,c;
int n,m,t;
void clear(data &a)
{
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)  a.a[i][j]=0;
}
void change(data &a,data b)
{
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	  a.a[i][j]=b.a[i][j];
}
data mul(data a,data b)
{
	data c; clear(c);
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	 {
	 	for (int k=1;k<=n;k++)
	 	 c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j]%p)%p;
	 }
	return c;
}
data pow(data num,int x)
{
	data ans; clear(ans);
	for (int i=1;i<=n;i++) ans.a[i][i]=1;
    data base; 
	change(base,num); 
	while (x)
	{
		if (x&1)  ans=mul(ans,base);
		x>>=1;
		base=mul(base,base);
	}
	return ans;
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d",&t);
	for (int T=1;T<=t;T++){
		scanf("%d%d",&n,&m);
		for (int i=1;i<=n;i++)
		 for (int j=1;j<=n;j++)  scanf("%d",&a.a[i][j]);
		data ans;  ans=pow(a,m);
		int sum=0;
		for (int i=1;i<=n;i++) sum=(sum+ans.a[i][i])%p;
		printf("%d\n",sum%p);
	}
}