首页 > 代码库 > HDU 4965 Fast Matrix Calculation(矩阵快速幂)
HDU 4965 Fast Matrix Calculation(矩阵快速幂)
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。
可以转化一下:(a*b)^(n*n)=a*(b*a)^(n*n-1)*b。b*a可以得到一个k*k的矩阵,k很小所以不会超时,快速幂一下就可以了啊。
Fast Matrix Calculation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 717 Accepted Submission(s): 355
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.
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.
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
#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <iomanip> #include <stdio.h> #include <string> #include <queue> #include <cmath> #include <stack> #include <map> #include <set> #define eps 1e-10 ///#define M 1000100 ///#define LL __int64 #define LL long long ///#define INF 0x7ffffff #define INF 0x3f3f3f3f #define PI 3.1415926535898 #define zero(x) ((fabs(x)<eps)?0:x) using namespace std; const int maxn = 1010; #define mod 6 struct matrix { int f[6][6]; }; int A[maxn][maxn]; int B[maxn][maxn]; int C[maxn][maxn]; int D[maxn][maxn]; matrix mul(matrix a, matrix b, int n) { matrix c; memset(c.f, 0, sizeof(c.f)); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { for(int k = 0; k < n; k++) c.f[i][j] += a.f[i][k]*b.f[k][j]; c.f[i][j] %= 6; } } return c; } matrix pow_mod(matrix a, int b, int n) { matrix s; memset(s.f, 0 , sizeof(s.f)); for(int i = 0; i < n; i++) s.f[i][i] = 1; while(b) { if(b&1) s = mul(s, a, n); a = mul(a, a, n); b >>= 1; } return s; } int main() { int n, k; while(cin >>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]); matrix c; memset(c.f, 0, sizeof(c.f)); for(int i = 0; i < k; i++) { for(int j = 0; j < k; j++) { for(int jj = 0; jj < n; jj++) c.f[i][j] += B[i][jj]*A[jj][j]; c.f[i][j] %= mod; } } c = pow_mod(c, n*n-1, k); memset(C, 0, sizeof(C)); memset(D, 0, sizeof(D)); for(int i = 0; i < n; i++) { for(int j = 0; j < k; j++) { for(int jj = 0; jj < k; jj++) C[i][j] += A[i][jj]*c.f[jj][j]; C[i][j] %= mod; } } int ans = 0; for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { for(int jj = 0; jj < k; jj++) D[i][j] += C[i][jj]*B[jj][j]; D[i][j] %= mod; ans += D[i][j]; } } cout<<ans<<endl; } return 0; }
HDU 4965 Fast Matrix Calculation(矩阵快速幂)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。