Tr A
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1769 Accepted Submission(s): 1302
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
Author
xhd
Source
HDU 2007-1 Programming Contest
Recommend
linle
#include<cstdio>
struct Matrix
{
int s[15][15];
};
Matrix A,B,D;
int n;
Matrix mul (Matrix A, Matrix B)
{
Matrix s;
int i, j, k;
for(i = 1; i <= n; i++)
for(j = 1; j <= n; j++)
{
s.s[i][j] = 0;
for(k = 1; k <= n; k++)
s.s[i][j] =(s.s[i][j] + A.s[i][k] * B.s[k][j]) % 9973;
}
return s;
}
//矩阵快速幂
Matrix pow (Matrix A, int k)
{
Matrix s = D;
while(k)
{
if(k & 1) s = mul(A, s);
A = mul(A,A);
k = k>>1;
}
return s;
}
int main ()
{
int T;
scanf("%d", &T);
while(T--)
{
int k;
int i, j;
scanf("%d %d", &n, &k);
for(i = 1; i <= n; i++)
for(j = 1; j <= n; j++)
{
D.s[i][j] = (i == j) ;
scanf("%d", &A.s[i][j]);
}
A = pow(A,k);
int sum = 0;
for(i = 1; i <= n; i++)
sum = (sum + A.s[i][i]) % 9973;
printf("%d\n", sum);
}
return 0;
}