HDOJ 4965 Fast Matrix Calculation

(AB)^n=A*(BA)^(n-1)^B

Fast Matrix Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 576    Accepted Submission(s): 297

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

HDOJ 4965 Fast Matrix Calculation,布布扣,bubuko.com

时间: 2024-10-08 10:08:57

HDOJ 4965 Fast Matrix Calculation的相关文章

HDU 4965 Fast Matrix Calculation 【矩阵】

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4965 题目大意:给你一个N*K的矩阵A以及一个K*N的矩阵B (4 <= N <= 1000)以及 (2 <=K <= 6),然后接下来四步: 算一个新的矩阵C=A*B 算M=C^ (N*N) 对于M中的每个元素%6 将M中每个元素加起来,算出和. 也就是求出A*B * A*B * A*B * A*B * A*B *--* A*B   但是A*B形成的矩阵是N*N,而N大小有可能是10

hdu 4965 Fast Matrix Calculation(矩阵快速幂)

题目链接:hdu 4965 Fast Matrix Calculation 题目大意:给定两个矩阵A,B,分别为N*K和K*N: 矩阵C = A*B 矩阵M=CN?N 将矩阵M中的所有元素取模6,得到新矩阵M' 计算矩阵M'中所有元素的和 解题思路:因为矩阵C为N*N的矩阵,N最大为1000,就算用快速幂也超时,但是因为C = A*B, 所以CN?N=ABAB-AB=AC′N?N?1B,C' = B*A, 为K*K的矩阵,K最大为6,完全可以接受. #include <cstdio> #inc

HDU 4965 Fast Matrix Calculation(矩阵快速幂)

HDU 4965 Fast Matrix Calculation 题目链接 矩阵相乘为AxBxAxB...乘nn次,可以变成Ax(BxAxBxA...)xB,中间乘n n - 1次,这样中间的矩阵一个只有6x6,就可以用矩阵快速幂搞了 代码: #include <cstdio> #include <cstring> const int N = 1005; const int M = 10; int n, m; int A[N][M], B[M][N], C[M][M], CC[N]

HDU 4965 Fast Matrix Calculation(矩阵高速幂)

HDU 4965 Fast Matrix Calculation 题目链接 矩阵相乘为AxBxAxB...乘nn次.能够变成Ax(BxAxBxA...)xB,中间乘n n - 1次,这样中间的矩阵一个仅仅有6x6.就能够用矩阵高速幂搞了 代码: #include <cstdio> #include <cstring> const int N = 1005; const int M = 10; int n, m; int A[N][M], B[M][N], C[M][M], CC[N

hdu 4965 Fast Matrix Calculation(矩阵快速幂)2014多校训练第9场

Fast Matrix Calculation                                                                   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description One day, Alice and Bob felt bored again, Bob knows Ali

HDU 4965 Fast Matrix Calculation (矩阵快速幂取模----矩阵相乘满足结合律)

http://acm.hdu.edu.cn/showproblem.php?pid=4965 利用相乘的可结合性先算B*A,得到6*6的矩阵,利用矩阵快速幂取模即可水过. 1 #include<iostream> 2 #include<stdio.h> 3 #include<iostream> 4 #include<stdio.h> 5 #define N 1010 6 #define M 1010 7 #define K 6 8 using namespa

HDU 4965 Fast Matrix Calculation(矩阵高速幂)

题目大意:给你两个数字n和k,然后给你两个矩阵a是n*k的和b是k*n的,矩阵c = a*b,让你求c^(n*n). 直接求的话c是n*n的矩阵所以是1000*1000.会超时的啊. 能够转化一下:(a*b)^(n*n)=a*(b*a)^(n*n-1)*b.b*a能够得到一个k*k的矩阵,k非常小所以不会超时.高速幂一下就能够了啊. Fast Matrix Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 13

hdu 4965 Fast Matrix Calculation

题目链接:hdu 4965,题目大意:给你一个 n*k 的矩阵 A 和一个 k*n 的矩阵 B,定义矩阵 C= A*B,然后矩阵 M= C^(n*n),矩阵中一切元素皆 mod 6,最后求出 M 中所有元素的和.题意很明确了,便赶紧敲了个矩阵快速幂的模板(因为编程的基本功不够还是调试了很久),然后提交后TLE了,改了下细节,加了各种特技,比如输入优化什么的,还是TLE,没办法,只好搜题解,看了别人的题解后才知道原来 A*B 已经是 n*n 的矩阵了,所以(A*B)n*n 的快速幂里的每个乘法都是

hdu 4965 Fast Matrix Calculation(矩阵快速幂)

题意: 给你一个N*K的矩阵A和一个K*N的矩阵B,设矩阵C=AB,M=C^(N*N),矩阵Mmod6后,所有数的和是多少 思路: 刚开始我是直接计算的,开了一个1000*1000的矩阵,结果直接爆掉了 后来看了看题解,发现想的很巧妙 观察 M=ABABAB....AB,AB每次都是1000*1000,可是BA确是6*6的 那么 M=A(BABABA..BA)B,我们让BA进行矩阵快速幂就不会爆掉了 M=A(BA)^(N*N-1)B,最后计算一下就好了 代码: #include <iostrea