Problem Description
One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her.
Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is not less than 0 and not greater than 5. Then he does similar thing again with a bit difference: he keeps throwing his dice for N times and each time repeat it for K times to write down a K*N matrix B, in which each element is not less than 0 and not greater than 5. With the two matrix A and B formed, Alice’s task is to perform the following 4-step calculation.
Step 1: Calculate a new N*N matrix C = A*B.
Step 2: Calculate M = C^(N*N).
Step 3: For each element x in M, calculate x % 6. All the remainders form a new matrix M’.
Step 4: Calculate the sum of all the elements in M’.
Bob just made this problem for kidding but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math.
Input
The input contains several test cases. Each test case starts with two integer N and K, indicating the numbers N and K described above. Then N lines follow, and each line has K integers between 0 and 5, representing matrix A. Then K lines follow, and each line has N integers between 0 and 5, representing matrix B.
The end of input is indicated by N = K = 0.
Output
For each case, output the sum of all the elements in M’ in a line.
Sample Input
4 2 5 5 4 4 5 4 0 0 4 2 5 5 1 3 1 5 6 3 1 2 3 0 3 0 2 3 4 4 3 2 2 5 5 0 5 0 3 4 5 1 1 0 5 3 2 3 3 2 3 1 5 4 5 2 0 0
Sample Output
14 56
Author
SYSU
Source
2014 Multi-University Training Contest 9
Recommend
We have carefully selected several similar problems for you: 5189 5188 5186 5185 5184
直接按题意的做法,搞出一个1000*1000的矩阵,光一次乘法就超时了
注意到C = A*B
C * C * ….C = A(B＊Ａ) (B*A) ….. *B
其中Ｂ＊Ａ是一个ｋ＊ｋ的矩阵，做n^2 - 1次的快速幂，这样就把时间控制下来了
/*************************************************************************
> File Name: hdu4965.cpp
> Author: ALex
> Mail: zchao1995@gmail.com
> Created Time: 2015年03月16日 星期一 17时51分56秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
int A[1010][10];
int B[10][1010];
int D[1010][1010];
int ans[1010][1010];
int K;
class MARTIX
{
public:
int mat[10][10];
MARTIX()
{
memset (mat, 0, sizeof(mat));
}
MARTIX operator * (const MARTIX &b)const;
MARTIX& operator = (const MARTIX &b);
};
MARTIX MARTIX :: operator * (const MARTIX &b)const
{
MARTIX C;
for (int i = 0; i < K; ++i)
{
for (int j = 0; j < K; ++j)
{
C.mat[i][j] = 0;
for (int k = 0; k < K; ++k)
{
C.mat[i][j] += this -> mat[i][k] * b.mat[k][j];
C.mat[i][j] %= 6;
}
}
}
return C;
}
MARTIX& MARTIX :: operator = (const MARTIX &b)
{
for (int i = 0; i < K; ++i)
{
for (int j = 0; j < K; ++j)
{
this -> mat[i][j] = b.mat[i][j];
}
}
return *this;
}
MARTIX fastpow(MARTIX A, int n)
{
MARTIX ans;
for (int i = 0; i < K; ++i)
{
ans.mat[i][i] = 1;
}
while (n)
{
if (n & 1)
{
ans = ans * A;
}
n >>= 1;
A = A * A;
}
return ans;
}
int main()
{
int n;
while (~scanf("%d%d", &n, &K))
{
if (!n && !K)
{
break;
}
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < K; ++j)
{
scanf("%d", &A[i][j]);
}
}
for (int i = 0; i < K; ++i)
{
for (int j = 0; j < n; ++j)
{
scanf("%d", &B[i][j]);
}
}
MARTIX C;
for (int i = 0; i < K; ++i)
{
for (int j = 0; j < K; ++j)
{
for (int k = 0; k < n; ++k)
{
C.mat[i][j] += B[i][k] * A[k][j];
C.mat[i][j] %= 6;
}
}
}
C = fastpow(C, n * n - 1);
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < K; ++j)
{
D[i][j] = 0;
for (int k = 0; k < K; ++k)
{
D[i][j] += A[i][k] * C.mat[k][j];
D[i][j] %= 6;
}
}
}
int res = 0;
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
{
ans[i][j] = 0;
for (int k = 0; k < K; ++k)
{
ans[i][j] += D[i][k] * B[k][j];
ans[i][j] %= 6;
}
res += ans[i][j];
}
}
printf("%d\n", res);
}
return 0;
}