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: 131072/131072 K (Java/Others)
Total Submission(s): 717 Accepted Submission(s): 355
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.
The end of input is indicated by N = K = 0.
#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; }